3 電場の導出

$$\vE\rt = -\Nabla\phi\rt -\frac{1}{c}\del{\vA\rt}{t} =-\Nabla\left(\frac{... ...right) -\frac{1}{c}\deL{t}\left(\frac{q \bm{\beta}(t')}{\kappa(t')R(t')}\right)$$

$$=q \frac{\Nabla\left\{\kappa(t')R(t')\right\}}{\kappa^2(t')R^2(t'... ... -\bm{\beta}(t')\del{\left\{ \kappa(t')R(t')\right\}}{t}}{c\kappa^2(t')R^2(t')}$$

$$= q \frac{ \vn(t')-\bm{\beta}(t')+\dfrac{\left(\vn(t')\cdot\bm{\b... ...(t')\cdot\dot{\bm{\beta}}(t')\right)}{c\kappa(t')}\vn(t')}{\kappa^2(t')R^2(t')}$$

$$\hspace{30mm} - q \frac{ R(t')\dot{\bm{\beta}}(t')-\left\{ -\dfra... ...bm{\beta}}(t')\right)}{\kappa(t')}\right\}\bm{\beta}(t')}{c\kappa^2(t')R^2(t')}$$

$$=q \frac{\kappa(t')\vn(t')-\kappa(t')\bm{\beta}(t')+\left(\vn(t')... ...beta}(t')\right)\bm{\beta}(t')+\beta^2(t')\bm{\beta}(t')}{\kappa^3(t')R^2 (t')}$$

$$\hspace{50mm} + q \frac{\left(\vn(t')\cdot\dot{\bm{\beta}}(t')\ri... ...left(\vn(t')\cdot\dot{\bm{\beta}}(t')\right)\bm{\beta}(t')}{c\kappa^3(t')R(t')}$$

$$=q\frac{\left(1-\beta^2(t')\right)\vn(t')-\left(1-\beta^2(t')\rig... ...left(\vn(t')\cdot\dot{\bm{\beta}}(t')\right)\bm{\beta}(t')}{c\kappa^3(t')R(t')}$$

$$= q\frac{\left(1-\beta^2(t')\right)\left(\vn(t')-\bm{\beta}(t')\r... ...)-\bm{\beta}(t')\right)\times \dot{\bm{\beta}}(t')\right\}}{c\kappa^3(t')R(t')}$$

$$=q \left[ \frac{\left(1-\beta^2\right)\left(\vn-\bm{\beta}\right)... ...\dot{\bm{\beta}}\right\}}{c\kappa^3 R} \right] =\vE_{{\rm vel}}+\vE_{{\rm rad}}$$ (27)

$$\qquad \because\,\, \vn(t')\times \left\{ \left(\vn(t')-\bm{\beta... ...}(t')\right)\vn(t')-\left(\vn(t')\cdot\dot{\bm{\beta}}(t')\right)\bm{\beta}(t')$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp