ランダウ・リフシッツ力学 §19 問題 2

ランダウ・リフシッツの力学(増補第3版) §19 問題 2 のメモです。

有効断面積の計算

\(U_1 = 0, U_2= -U_0\)を問題7の表式に代入すると \[ \cfrac{\sin \alpha}{\sin \beta} = \sqrt{1 + \cfrac{2U_0}{mv_\infty}} \] \(\chi\)と\(\alpha, \beta\)の関係はまず \begin{align} (\varphi_0 -\alpha) + \beta & = \cfrac{\pi}{2} \\ 2 \varphi_0 + 2 (\beta – \alpha ) & = \pi \\ 2 \varphi_0 & = 2 (\alpha – \beta ) + \pi \\ \end{align} この関係を\(\chi = 2\varphi_0 – \pi \)に代入して \[ \chi = 2\varphi_0 – \pi = 2(\alpha – \beta). \] 次に \[ \cfrac{\sin \beta}{\sin \alpha} = \cfrac{1}{n} \] に上の関係を代入すると \begin{align} \cfrac{\sin \left(\alpha – \cfrac{\chi}{2} \right)}{\sin \alpha} & = \cfrac{\sin \alpha \cos \left(\cfrac{\chi}{2}\right) – \cos \alpha \sin\left(\cfrac{\chi}{2}\right) } {\sin \alpha} \\ & = \cos \left(\cfrac{\chi}{2}\right) – \cot \alpha \sin\left(\cfrac{\chi}{2}\right) = \cfrac{1}{n} \\ \end{align} 書き直して \[ \cot \alpha = \cfrac{\cos \cfrac{\chi}{2} – \cfrac{1}{n}}{\sin \cfrac{\chi}{2}} \] これと \[ a \sin \alpha = \rho \] との関係から\(\alpha\)を消去します: \[ \cot^2 \alpha = \cfrac{1}{\sin^2 \alpha} -1 = \cfrac{a^2}{\rho^2} – 1 \] \begin{align} \cfrac{a^2}{\rho ^2} -1 = & \cfrac{\left( \cos \cfrac{\chi}{2} -\cfrac{1}{n}\right)}{\sin^2 \cfrac{\chi}{2}} \\ \cfrac{a^2}{\rho ^2} = & 1 + \cfrac{\left( \cos \cfrac{\chi}{2} -\cfrac{1}{n}\right)}{\sin^2 \cfrac{\chi}{2}} \\ = & \cfrac{\sin^2\cfrac{\chi}{2} \left( \cos \cfrac{\chi}{2} -\cfrac{1}{n}\right)}{\sin^2 \cfrac{\chi}{2}} \\ = & \cfrac{ 1 – \cfrac{2}{n} \cos \cfrac{\chi}{2} + \cfrac{1}{n^2} }{\sin^2 \cfrac{\chi}{2}} \\ = & \cfrac{ 1 – 2n \cos \cfrac{\chi}{2} + n^2 }{n^2 \sin^2 \cfrac{\chi}{2}} \\ \end{align} これから \[ \rho^2 = a^2 \cfrac{n^2 \sin^2 \cfrac{\chi}{2}}{ 1 – 2n \cos \cfrac{\chi}{2} + n^2 } . \] \begin{align} \cfrac{d\rho^2}{d \chi} =& 2\rho \cfrac{d \rho}{d \chi} \\ = & a^2 n^2 \left[\cfrac{ 2 \sin \cfrac{\chi}{2} \cos \cfrac{\chi}{2} \cfrac{1}{2}}{ 1 – 2n \cos \cfrac{\chi}{2} + n^2 } + \cfrac{\sin^2 \cfrac{\chi}{2} (-1)(-2n)(-\sin \cfrac{\chi}{2})\cfrac{1}{2} }{ (1 – 2n \cos \cfrac{\chi}{2} + n^2)^2 } \right] \\ = & a^2 n^2 \sin \cfrac{\chi}{2}\left[\cfrac{ \cos \cfrac{\chi}{2} }{ 1 – 2n \cos \cfrac{\chi}{2} + n^2 } – \cfrac{n \sin^2 \cfrac{\chi}{2} }{ (1 – 2n \cos \cfrac{\chi}{2} + n^2)^2 } \right] \\ = & a^2 n^2 \cfrac{\sin \cfrac{\chi}{2}}{(1 – 2n \cos \cfrac{\chi}{2} + n^2)^2} \left[ \cos \cfrac{\chi}{2} (1 – 2n \cos \cfrac{\chi}{2} + n^2 ) – n \sin^2 \cfrac{\chi}{2} \right] \\ = & a^2 n^2 \cfrac{\sin \cfrac{\chi}{2}}{(1 – 2n \cos \cfrac{\chi}{2} + n^2)^2} \left[ (1+n^2)\cos \cfrac{\chi}{2} – 2n \cos^2 \cfrac{\chi}{2} – n (1 – \cos^2 \cfrac{\chi}{2}) \right] \\ = & a^2 n^2 \cfrac{\sin \cfrac{\chi}{2}}{(1 – 2n \cos \cfrac{\chi}{2} + n^2)^2} \left[ (1+n^2)\cos \cfrac{\chi}{2} – n \cos^2 \cfrac{\chi}{2} – n \right] \\ = & a^2 n^2 \cfrac{\sin \cfrac{\chi}{2}}{\left(1 – 2n \cos \cfrac{\chi}{2} + n^2\right)^2} \left[ n \left(n\cos \cfrac{\chi}{2} -1 \right) – \cos^2 \cfrac{\chi}{2}\left(n \cos \cfrac{\chi}{2} – 1\right) \right] \\ = & a^2 n^2 \cfrac{\sin \cfrac{\chi}{2}}{\left(1 – 2n \cos \cfrac{\chi}{2} + n^2\right)^2} \left[ \left(n\cos \cfrac{\chi}{2} -1 \right) \left(n – \cos \cfrac{\chi}{2} \right) \right] \\ \end{align} 最終的に散乱有効断面積は \begin{align} d\sigma = & \cfrac{\rho(\chi)}{\sin \chi}\left| \cfrac{d\rho}{d\chi}\right| do \\ = & \cfrac{\rho(\chi)}{2\sin \cfrac{\chi }{2} \cos \cfrac{\chi }{2}}\left| \cfrac{d\rho}{d\chi}\right| do \\ = & \cfrac{a^2 n^2}{4 \cos \cfrac{\chi}{2}\left(1 – 2n \cos \cfrac{\chi}{2} + n^2\right)^2} \left[ \left(n\cos \cfrac{\chi}{2} -1 \right) \left(n – \cos \cfrac{\chi}{2} \right) \right] do .\\ \end{align}

全有効断面積の計算

全有効断面積の計算をします:

\[\cos\cfrac{\chi}{2} = x\]と置換すると積分範囲は\(x=1\)から\(x=1/n\)になります。また \[ -\sin\cfrac{\chi}{2} \cfrac{d\chi}{2} = dx \] \[ -\sin\cfrac{\chi}{2} d\chi = 2dx \] あるいは \begin{align} \int d\sigma = & \int_0^{\chi_\mathrm{max}} \cfrac{a^2 n^2}{4 \cos \cfrac{\chi}{2}\left(1 – 2n \cos \cfrac{\chi}{2} + n^2\right)^2} \left[ \left(n\cos \cfrac{\chi}{2} -1 \right) \left(n – \cos \cfrac{\chi}{2} \right) \right] 2\pi \sin \chi d\chi \\ = & \int_{0}^{\chi_\mathrm{max}} \cfrac{a^2 n^2}{4 \cos \cfrac{\chi}{2}\left(1 – 2n \cos \cfrac{\chi}{2} + n^2\right)^2} \left[ \left(n\cos \cfrac{\chi}{2} -1 \right) \left(n – \cos \cfrac{\chi}{2} \right) \right] 4\pi \sin \cfrac{\chi}{2}\cos \cfrac{\chi}{2} d\chi \\ = & – \int_{1}^{1/n} \cfrac{a^2 n^2 \pi }{ \left(1 – 2n x + n^2\right)^2} \left[ \left(nx -1 \right) \left(n – x \right) \right] 2 dx \\ = & 2 a^2 n^2 \pi \int^{1}_{1/n} \cfrac{\left(nx -1 \right) \left(n – x \right) }{ \left(1 – 2n x + n^2\right)^2} dx \\ \end{align}

ここでさらに変数を\(1+n^2-2nx =u\)と変換します。こうすると\(dx = -du/2n\)

\begin{align} x = & \cfrac{1+n^2-u}{2n} \\ nx-1 = & \cfrac{n^2-1-u}{2} = -\cfrac{u – n^2+1}{2}\\ n -x = & \cfrac{n^2-1 + u}{2n} \end{align}

積分範囲は\(n^2-1\)から\((n-1)^2\)となります。

\begin{align} \int d\sigma = & 2 a^2 n^2 \pi \int^{1}_{1/n} \cfrac{\left(nx -1 \right) \left(n – x \right) }{ \left(1 – 2n x + n^2\right)^2} dx \\ = & 2 a^2 n^2 \pi \int^{(n-1)^2}_{n^2-1} \cfrac{( u+n^2-1) (u-n^2+1) }{ -4n u^2 } \cfrac{du}{-2n} \\ = & \cfrac{a^2 n^2 \pi}{4n^2} \int^{(n-1)^2}_{n^2-1} \cfrac{ u^2 – (n^2-1)^2 }{ u^2 } du\\ = & \cfrac{a^2 \pi}{4} \left[ u + \cfrac{(n^2-1)^2}{u} \right]^{(n-1)^2}_{n^2-1} \\ = & \cfrac{a^2 \pi}{4} \left[ (n-1)^2 – (n^2 -1) + (n^2-1)^2 \left( \cfrac{1}{(n-1)^2} – \cfrac{1}{n^2-1} \right) \right] \\ = & \cfrac{a^2 \pi}{4} \left[ (n-1)^2 – (n^2 -1) + (n+1)^2 – (n^2-1) \right] \\ = & \cfrac{a^2 \pi}{4} \left[ (n-1)^2 + (n+1)^2 – 2(n^2-1) \right] \\ = & \cfrac{a^2 \pi}{4} \left[ n^2 – 2n +1 + n^2 + 2n + 1 – 2n^2 + 2 \right] \\ = & \cfrac{a^2 \pi}{4} 4 \\ = & a^2 \pi . \\ \end{align}

となります。最終的に\(\pi a^2\)になることが確認できました。