ランダウ・リフシッツ力学 §15 問題 1

ランダウ・リフシッツの力学(増補第3版) §15 問題 1 のメモです。 \(E=0\), \(U = -\alpha/r\)を代入すると \begin{align} t & = \int \cfrac {dr}{\sqrt{\cfrac{2}{m} (E-U) – \cfrac{M^2}{m^2r^2} }} \\ & = \int \cfrac {dr}{\sqrt{\cfrac{2\alpha}{mr} – \cfrac{M^2}{m^2r^2} }} \\ & = \int \cfrac {rdr}{\sqrt{\cfrac{2\alpha r}{m} – \cfrac{M^2}{m^2} }} \end{align} \[ r = \cfrac{M^2}{2m\alpha} (1 +\eta^2) = \cfrac{p}{2}(1 + \eta^2) \] を代入すると \[ dr = \cfrac{p}{2} 2\eta d\eta = p\eta d\eta \] を使って \begin{align} t & = \int \cfrac {rdr}{\sqrt{\cfrac{2\alpha r}{m} – \cfrac{M^2}{m^2} }} \\ = \int \cfrac {\cfrac{p}{2}(1 + \eta^2) p\eta d\eta} {\sqrt{\cfrac{2\alpha }{m}\cfrac{M^2}{2m\alpha}(1 + \eta^2) – \cfrac{M^2}{m^2} }} \\ = \int \cfrac {\cfrac{p}{2}(1 + \eta^2) p\eta d\eta} { \cfrac{M}{m}\eta } \\ = \cfrac{m}{M} \cfrac{p^2}{2}\left(\eta + \cfrac{1}{3}\eta^3 \right) \\ = \cfrac{m}{\sqrt{pm\alpha}} \cfrac{p^2}{2}\left(\eta + \cfrac{1}{3}\eta^3 \right) \\ = \sqrt{\cfrac{mp^3}{ \alpha }} \cfrac{\eta}{2}\left(1 + \cfrac{1}{3}\eta^2 \right) \\ \end{align} \(E=0\)ですから\(e=1\)です: \[ ex = x= er\cos\varphi = p-r = p – \cfrac{p}{2}(1 + \eta^2)= \cfrac{p}{2}(1 – \eta^2) \] \[ y = \sqrt{r^2-x^2} = \sqrt{\cfrac{p^2}{4}(1+\eta^2)^2 – \cfrac{p^2}{4}(1-\eta^2)^2 } = \sqrt{\cfrac{p^2}{4} 2 \eta^2 2} = p\eta \]