ランダウ・リフシッツ力学 §15 問題 2

ランダウ・リフシッツの力学(増補第3版) §15 問題 2 のメモです。

\(t\)

\begin{align} t = & \int \cfrac{dr}{\sqrt{\cfrac{2}{m}\left( E + \cfrac{\alpha}{r^2}\right) – \cfrac{M^2}{m^2 r^2} }}\\ = & \sqrt{\cfrac{m}{2}} \int \cfrac{r dr}{\sqrt{ E r^2 + \alpha – \cfrac{M^2}{2 m} }}\\ \end{align}

\( Er^2 = u\)と置くと

\[ rdr = \cfrac{du}{2E} \]

これから

\begin{align} t = & \cfrac{1}{2E}\sqrt{\cfrac{m}{2}} \int \cfrac{du}{\sqrt{ u + \alpha – \cfrac{M^2}{2 m} }}\\ = & \cfrac{1}{2E}\sqrt{\cfrac{m}{2}} \times 2 \sqrt{ u + \alpha – \cfrac{M^2}{2 m} }\\ = & \cfrac{1}{E}\sqrt{\cfrac{m}{2}} \sqrt{ Er^2 + \alpha – \cfrac{M^2}{2 m} }\\ \end{align}

\(\varphi \)

\(M/r = u\) \[ -\cfrac{M}{r^2} dr = du \]

(14.7)に\(U = -\alpha/r\)を代入して

\begin{align} \varphi = & \int \cfrac{\cfrac{M}{r^2} dr}{\sqrt{2m \left( E + \cfrac{\alpha}{r^2}\right) – \cfrac{M^2}{ r^2} }}\\ = & \int \cfrac{- du}{\sqrt{ 2mE + \cfrac{2m \alpha u^2 }{M^2} – u^2 }}\\ = & \int \cfrac{- du}{\sqrt{ 2mE + \left( \cfrac{2m \alpha }{M^2} – 1 \right) u^2 }}\\ \end{align}

\(E>0\) , \(1 > 2m\alpha / M^2\)の場合

\begin{align} \varphi = & \cfrac{1}{\sqrt{1 – \cfrac{2m\alpha}{M^2}}} \int \cfrac{- du}{\sqrt{ \cfrac{2mE}{ 1 – \cfrac{2m\alpha}{M^2} } – u^2 }}\\ = & \cfrac{1}{\sqrt{1 – \cfrac{2m\alpha}{M^2}}} \cos^{-1} \cfrac{u}{\sqrt{ \cfrac{2mE}{ 1 – \cfrac{2m\alpha}{M^2} } }}\\ \cos \left( \varphi \sqrt{1 – \cfrac{2m\alpha}{M^2}} \right) = & \cfrac{u}{\sqrt{ \cfrac{2mE}{ 1 – \cfrac{2m\alpha}{M^2} } }}\\ = & \cfrac{M}{r \sqrt{ \cfrac{2mE}{ 1 – \cfrac{2m\alpha}{M^2} } }}\\ = & \cfrac{1}{r \sqrt{ \cfrac{2mE}{ M^2 – 2m\alpha } }}\\ \end{align}

\(E>0\) , \(1 < 2m\alpha / M^2\)の場合

\begin{align} \varphi = & \cfrac{1}{\sqrt{ \cfrac{2m\alpha}{M^2} -1 }} \int \cfrac{- du}{\sqrt{ \cfrac{2mE}{ \cfrac{2m\alpha}{M^2} -1 } + u^2 }}\\ = & \cfrac{1}{\sqrt{ \cfrac{2m\alpha}{M^2} – 1 }} \sinh^{-1} \cfrac{u}{\sqrt{ \cfrac{2mE}{ \cfrac{2m\alpha}{M^2} -1 } }}\\ \sinh \left( \varphi \sqrt{ \cfrac{2m\alpha}{M^2} -1 } \right) = & \cfrac{u}{\sqrt{ \cfrac{2mE}{ \cfrac{2m\alpha}{M^2} -1 } }}\\ = & \cfrac{M}{r \sqrt{ \cfrac{2mE}{ \cfrac{2m\alpha}{M^2} – 1 } }}\\ = & \cfrac{1}{r \sqrt{ \cfrac{2mE}{ 2m\alpha – M^2 } }}\\ \end{align}

\(E< 0\) , \(1 < 2m\alpha / M^2\)の場合

\(E< 0\)の時は、根号の中が正になるのは\(1 < 2m\alpha / M^2\)の場合だけです。

\begin{align} \varphi = & \cfrac{1}{\sqrt{ \cfrac{2m\alpha}{M^2} -1 }} \int \cfrac{- du}{\sqrt{ – \cfrac{2m|E|}{ \cfrac{2m\alpha}{M^2} -1 } + u^2 }}\\ = & \cfrac{1}{\sqrt{ \cfrac{2m\alpha}{M^2} – 1 }} \cosh^{-1} \cfrac{u}{\sqrt{ \cfrac{2m|E|}{ \cfrac{2m\alpha}{M^2} -1 } }}\\ \cosh \left( \varphi \sqrt{ \cfrac{2m\alpha}{M^2} -1 } \right) = & \cfrac{u}{\sqrt{ \cfrac{2m|E|}{ \cfrac{2m\alpha}{M^2} -1 } }}\\ = & \cfrac{M}{r \sqrt{ \cfrac{2m|E|}{ \cfrac{2m\alpha}{M^2} – 1 } }}\\ = & \cfrac{1}{r \sqrt{ \cfrac{2m|E|}{ 2m\alpha – M^2 } }}\\ \end{align}