ランダウ・リフシッツ力学 §15 問題 3

ランダウ・リフシッツの力学(増補第3版) §15 問題 3 のメモです。

式(14.10)を書き直します:

$\begin{align} \Delta \varphi & = 2 \int_{r_\mathrm{min}}^{r_\mathrm{max}} \cfrac{\cfrac{M}{r^2} dr } {\sqrt{2m\left(E-U \right) -\cfrac{M^2}{r^2} }} \\ & = – 2 \cfrac{\partial}{\partial M}\int_{r_\mathrm{min}}^{r_\mathrm{max}} \sqrt{2m\left(E-U \right) -\cfrac{M^2}{r^2} } dr \end{align}$ 根号の中を書き直します:

$\begin{align} \sqrt{2m\left(E+ \cfrac{\alpha}{r} – \delta U \right) -\cfrac{M^2}{r^2} } & = \sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} } \sqrt{1 – \cfrac{2m\delta U}{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2}}} \\ & \approx \sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} } \left( 1 – \cfrac{1}{2} \cfrac{2m\delta U}{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2}} \right) \\ & = \sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} } – \cfrac{1}{2} \cfrac{2m\delta U}{\sqrt{ 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} \\ \end{align}$ これから

$\Delta \varphi$ の積分は次のようになります:

$\Delta \varphi = – 2 \cfrac{\partial}{\partial M}\int_{r_\mathrm{min}}^{r_\mathrm{max}} \left[ \sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} } – \cfrac{1}{2} \cfrac{2m\delta U}{\sqrt{ 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} \right] dr$

0次項

$\delta U$ について0次項を

$\Delta \varphi_0$ とすると(この項は本文で既に計算済みですので、その結果を用います)

$\begin{align} \Delta \varphi_0 & = – 2 \cfrac{\partial}{\partial M} \int_{r_\mathrm{min}}^{r_\mathrm{max}} \sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} } dr \\ & = 2 \int_{r_\mathrm{min}}^{r_\mathrm{max}} \cfrac{\cfrac{M}{r^2}dr} {\sqrt{2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} \\ & = 2 \left. \cos^{-1} \cfrac{\cfrac{M}{r} – \cfrac{m\alpha}{M}} {\sqrt{2mE+\cfrac{m^2 \alpha^2}{M^2} }} \right|_{r_\mathrm{min}}^{r_\mathrm{max}}　\\ & = 2 \left. \cos^{-1} \cfrac{\cfrac{M}{r} – \cfrac{m\alpha}{M}} {\cfrac{m\alpha}{M} \sqrt{1 +\cfrac{2mE }{m^2 \alpha^2} }} \right|_{r_\mathrm{min}}^{r_\mathrm{max}}　\\ & = 2 \left[ \cos^{-1} \cfrac{\cfrac{M}{r_\mathrm{max}} – \cfrac{m\alpha}{M}} {e \cfrac{m\alpha}{M} } – \cos^{-1} \cfrac{\cfrac{M}{r_\mathrm{min}} – \cfrac{m\alpha}{M}} {e \cfrac{m\alpha}{M} } \right] \\ & = 2 \left[ \cos^{-1} \cfrac{\cfrac{Mm\alpha (1-e)}{M^2 } – \cfrac{m\alpha}{M}} {e \cfrac{m\alpha}{M} } – \cos^{-1} \cfrac{\cfrac{Mm\alpha (1+e) }{M^2 } – \cfrac{m\alpha}{M}} {e \cfrac{m\alpha}{M} } \right] \\ & = 2 \left[ \cos^{-1} (-1) – \cos^{-1}(1) \right] \\ & = 2\pi \end{align}$

$r_\mathrm{min} = p/(1-e^2)$ ,

$r_\mathrm{max} = p/(1+e^2)$ を用いました。

1次項

$\begin{align} \delta \varphi & = – 2 \cfrac{\partial}{\partial M}\int_{r_\mathrm{min}}^{r_\mathrm{max}} \left[ – \cfrac{1}{2} \cfrac{2m\delta U}{\sqrt{ 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} \right] dr \\ & = \cfrac{\partial}{\partial M}\int_{r_\mathrm{min}}^{r_\mathrm{max}} \cfrac{2m\delta U}{\sqrt{ 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} dr \end{align}$

$p/r = 1 + e\cos\varphi$ の関係を使って根号の中を書き直します:

$\begin{align} 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} & = 2mE+ \cfrac{2m \alpha }{p}(1+ e\cos\varphi) -\cfrac{M^2}{p^2}(1+e\cos\varphi)^2 \\ & = 2mE – \cfrac{m^2 \alpha^2 }{M^2} \left[ (1+ e\cos\varphi) -1 \right]^2 +\cfrac{m^2\alpha^2}{M^2} \\ & = 2mE – \cfrac{m^2 \alpha^2 }{M^2} e^2\cos^2\varphi +\cfrac{m^2\alpha^2}{M^2} \\ & = 2mE – \cfrac{m^2 \alpha^2 }{M^2} \left( 1 + \cfrac{2EM^2}{m\alpha^2}\right)\cos^2\varphi +\cfrac{m^2\alpha^2}{M^2} \\ & = \left( 2mE + \cfrac{m^2 \alpha^2 }{M^2}\right)\sin^2\varphi \\ & = \cfrac{m^2\alpha^2}{M^2}\left( 1 + \cfrac{2EM^2 }{m\alpha^2}\right)\sin^2\varphi \\ & = \cfrac{M^2e^2}{p^2} \sin^2\varphi \\ \end{align}$ 一方

$dr$ は

$\begin{align} -\cfrac{pdr}{r^2} & = – e \sin\varphi d\varphi \\ dr & = \cfrac{e}{p} r^2 \sin\varphi d\varphi \\ \end{align}$ となります。これらを用いて

$\begin{align} \delta \varphi & = \cfrac{\partial}{\partial M}\int_{r_\mathrm{min}}^{r_\mathrm{max}} \cfrac{2m\delta U}{\sqrt{ 2m\left(E+ \cfrac{\alpha}{r} \right) -\cfrac{M^2}{r^2} }} dr\\ & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m\delta U}{\sqrt{\cfrac{M^2e^2}{p^2} \sin^2\varphi }} \cfrac{e}{p} r^2 \sin\varphi d\varphi \\ & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m\delta U}{ M } r^2 d\varphi \\ \end{align}$

a)

$\begin{align} \delta \varphi & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m \beta}{ M r^2 } r^2 d\varphi \\ & = \cfrac{\partial}{\partial M} \cfrac{2m \pi \beta}{ M } & = – \cfrac{2m \pi \beta}{ M^2 } & = – \cfrac{2\pi \beta }{ p \alpha } \end{align}$

b)

$\begin{align} \delta \varphi & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m \gamma}{ M r^3 } r^2 d\varphi \\ & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m \gamma}{ M r } d\varphi \\ & = \cfrac{\partial}{\partial M}\int_{0}^{\pi} \cfrac{2m \gamma}{ Mp }(1+e\cos\varphi) d\varphi \\ & = \cfrac{\partial}{\partial M} \cfrac{2m^2 \pi \alpha\gamma }{ M^3} \\ & = – \cfrac{6m^2 \pi \alpha\gamma }{ M^4} \\ & = – \cfrac{6m^2 \pi \alpha\gamma }{ p^2 m^2 \alpha^2} \\ & = – \cfrac{6 \pi \gamma }{ p^2 \alpha} \\ \end{align}$