5-3-(b)

$\displaystyle \int \left[\kappa g^2\right]\,d\Omega$	$\displaystyle = 2\pi \left[ \dot{\beta}^2 \left\{ I_3 + 2\beta J_4 -\gamma^{-2}... ...\frac{1}{2\gamma^2} \left\{ I_5-3K_5 \right\} \right) \sin^2 i \right\} \right]$
	$\displaystyle =\frac{8\pi}{3c^2} \left[ \dot{v}^2 \gamma^6 \left( 1-\beta^2\sin^2 i\right) \right]$	(23)

5-3-(b)解答

$\displaystyle \int\cos\phi\,d\Omega= \int_0^{\pi}\sin\theta d\theta\int_0^{2\pi}\cos\phi d\phi =\int_0^{\pi}\sin\theta\,d\theta \Big[\sin\phi\Big]_0^{2\pi} =0$

$\displaystyle \sin^2 i\, \sin^2\theta\,\cos^2\phi+\cos^2 i\,\cos^2\theta =\sin^2 i \left[ \cos^2\phi -\left(1+\cos^2\phi\right)\cos^2\theta\right]+\cos^2\theta$

$\displaystyle \int \left[\kappa g^2\right]\,d\Omega$	$\displaystyle = \dot{\beta}^2 \Bigg[ \int d\Omega \frac{1}{\kappa^3} + 2\beta(1... ... \frac{\cos^2 \phi -\left(1+\cos^2\phi\right) \cos^2\theta }{\kappa^5}\,d\Omega$
	$\displaystyle \hspace{30mm} -\gamma^{-2}\int \frac{\cos^2\theta}{\kappa^5}d\Ome... ...\beta}{\kappa^4}-\frac{\cos\theta}{\gamma^2 \kappa^5}\right)d\Omega}_{=0}\Bigg]$
	$\displaystyle =\dot{\beta}^2\left[ 2\pi I_3 +4\pi \beta (1-\sin^2 i)J_4 -\sin^2 i \, \gamma^{-2} \left( \pi I_5 -3\pi K_5 \right) \right]$
	$\displaystyle =2\pi \left[ \dot{\beta}^2 \left( I_3 + 2\beta J_4 -\gamma^{-2} K... ...J_4 +\frac{1}{2\gamma^2} \left(I_5 - 3K_5\right)\right) \sin^2 i\right) \right]$

$\displaystyle I_3$	$\displaystyle = \frac{\left(1+\beta\right)^2 -\left(1-\beta\right)^2}{2\beta \left(1-\beta^2\right)^2} = \frac{2}{\left(1-\beta^2\right)^2} ＝2 \gamma^4$
$\displaystyle I_5$	$\displaystyle = \frac{\left(1+\beta\right)^4 -\left(1-\beta\right)^4}{4\beta \l... ...ight)^4} = 2 \frac{1+\beta^2}{\left(1-\beta^2\right)^4} = 2(1+\beta^2) \gamma^8$
$\displaystyle J_4$	$\displaystyle = \frac{1}{3} \di{I_3}{\beta} = \frac{1}{3} \left[ 2(-2) (-2 \bet... ... = \frac{8}{3} \frac{\beta}{\left(1-\beta^2\right)^3} =\frac{8}{3}\beta\gamma^6$
$\displaystyle K_5$	$\displaystyle = \frac{1}{4} \di{J_4}{\beta} = \frac{2}{3} \frac{\left(1-\beta^2... ...\frac{5\beta^2 +1 }{\left(1-\beta^2\right)^4} =\frac{2}{3}(5\beta^2 +1)\gamma^8$

$\displaystyle I_3 + 2\beta J_4 -\gamma^{-2} K_5$	$\displaystyle = 2\gamma^4 +2\beta \frac{8}{3} \beta \gamma^6 -\gamma^{-2} \frac... ...ac{2}{\gamma^2} + \frac{16}{3} \beta^2 -\frac{10}{3}\beta^2 -\frac{2}{3}\right)$
	$\displaystyle =\gamma^6 \, \frac{4}{3}$
$\displaystyle 2\beta J_4 + \frac{\gamma^{-2}}{2} I_5 -\frac{3}{2} \gamma^{-2} K_5$	$\displaystyle = 2\beta \frac{8}{3} \beta \gamma^6 +\frac{\gamma^{-2}}{2}2\left(... ...mma^6 \left[ \frac{16}{3} \beta^2 + \left(1+\beta^2\right) -5\beta ^2 -1\right]$
	$\displaystyle = \gamma^6 \, \frac{4}{3}\beta^2$

5-3-(b)

5-3-(b)解答

脚注