2-a

以下の関係式を示せ。

$$\vn\cdot\bm{\beta} = \beta \cos\theta, \quad \vn\cdot \dot{\bm{\b... ...ta\cos i\right), \quad \dot{\bm{\beta}}\cdot \bm{\beta} =\dot{\beta}\beta\cos i$$ (13)

$$\kappa g^2 = \dot{\beta}^2\Bigg\{ \frac{1}{\kappa^3} +\frac{2}{\kappa^4}\beta \left(\cos i \, \sin i \,\sin\theta \,\cos\phi +\cos^2 i\, \cos\theta\right)$$

$$\hspace{30mm} -\frac{1}{\kappa^5}\gamma^{-2}\left( \sin^2 i \,\si... ...\,\sin\theta\, \cos\theta \,\cos\phi + \cos^2 i\, \cos^2 \theta \right\} \Bigg)$$ (14)

$$I_{n+1} \equiv \int_{-1}^{+1} \frac{d\mu}{\left(1-\beta\mu\right)^{n+1}} ... ...\left(1+\beta\right)^n-\left(1-\beta\right)^n}{n\beta \left(1-\beta^2\right)^n}$$ (15)

$$J_{n+1} \equiv \int_{-1}^{+1} \frac{\mu d\mu}{\left(1-\beta\mu\right)^{n+1}} = \frac{1}{n}\di{I_n}{\beta}$$ (16)

$$K_{n+1} \equiv \int_{-1}^{+1} \frac{\mu^2d\mu}{\left(1-\beta\mu\right)^{n+1}} = \frac{1}{n}\di{J_n}{\beta}$$ (17)

2-a解答

図より、

$$\bm{\beta}=\beta\left(\,0,\,0,\,1\right),\quad \dot{\bm{\beta}} = \dot{\beta}\left( \,\sin i,\, 0,\, \cos i\right)$$

であり、又 $\vv=c\bm{\beta}$であるから、

$$\vn \cdot \bm{\beta} = \beta \cos\theta$$

$$\vn \cdot \dot{\bm{\beta}} = \dot{\beta} \left(\sin\theta\,\cos\phi\,\sin i + \cos\theta\, \cos i\right)$$

$$\dot{\bm{\beta}}\cdot \bm{\beta} = \dot{\beta}\beta \cos i$$

を得る。この関係より

$$\kappa g^2 = \kappa \frac{1}{\kappa^6} \left[ \vn \times \left\{\left(\vn -\... ...m{\beta}\right) - \left(1-\vn \cdot \bm{\beta}\right)\dot{\bm{\beta}}\right\}^2$$

$$=\frac{1}{\kappa^5} \left[ \left(\vn \cdot \dot{\bm{\beta}}\right... ...left(\vn \cdot \dot{\bm{\beta}} -\bm{\beta}\cdot \dot{\bm{\beta}}\right)\right]$$

$$= \frac{1}{\kappa^5} \Bigg[ \dot{\beta}^2 \left(\sin^2\theta\,\co... ...os\theta\, \sin i\, \cos i \, \cos\phi\right) \left(\beta^2 -1 + 2\kappa\right)$$

$$\hspace{30mm} + \kappa^2 \dot{\beta}^2 -2 \dot{\beta} \left(\sin\... ...sin\theta\,\cos\phi \, \sin i + \cos\theta\, \cos i -\beta \cos i\right) \Bigg]$$

$$= \frac{\dot{\beta}^2}{\kappa^3} + \frac{2\dot{\beta}^2\beta}{\kappa^4} \left( \cos i\,\sin\theta\,\cos\phi\,\sin i + \cos\theta \,\cos^2 i \right)$$

$$\hspace{30mm} + \dot{\beta}^2 \frac{\left(\beta^2-1\right)}{\kapp... ... \cos i\, \sin\theta\, \cos\theta\, \cos\phi + \cos^2 i \,\cos^2 \theta \right)$$

$$= \dot{\beta}^2\Bigg\{ \frac{1}{\kappa^3} +\frac{2}{\kappa^4}\beta \left(\cos i \, \sin i \,\sin\theta \,\cos\phi +\cos^2 i\, \cos\theta\right)$$

$$\hspace{30mm} -\frac{1}{\kappa^5}\gamma^{-2} \left( \sin^2 i\, \s... ... i \, \sin\theta\,\cos\theta\,\cos\phi + \cos^2i \, \cos^2\theta\right) \Bigg\}$$

を得る。次にEq.(15),(16),(17)は

$$I_{n+1} = \int_{-1}^{+1} \frac{d\mu}{\left(1-\beta\mu\right)^{n+1}} ; \qq... ...cc} \mu & -1 & \to & +1 \\ \hline x & 1+\beta\mu & \to & 1-\beta\mu \end{array}$$

$$= - \int_{1+\beta}^{1-\beta} \frac{x^{-(n+1)}}{\beta} dx = \frac{... ...left(1+\beta\right)^n -\left(1-\beta\right)^n}{n\beta \left(1-\beta^2\right)^n}$$

$$\di{I_n}{\beta} = \dI{\beta} \int_{-1}^{+1} \frac{d\m... ...eta \mu\right)^{n+1}},\quad \therefore \, J_{n+1} = \frac{1}{n} \di{I_n}{\beta}$$

$$\di{J_n}{\beta} = \dI{\beta} \int_{-1}^{+1} \frac{\mu... ...ta \mu\right)^{n+1}}, \quad \therefore \, K_{n+1} = \frac{1}{n} \di{J_n}{\beta}$$

となるので、諸関係式が導けた。 著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp