3 ベクトルポテンシャルの時間微分

$\vA \xt$ を $t$ で偏微分すると、

$$\deL{t} \vA\xt = \frac{q\mu_0}{4\pi} \Int dt' \, \deL{t} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right) \frac{\vu(t')}{R(\vx,t')}\right\}$$

$$= -\frac{q}{4\pi \vepsilon_0}\frac{1}{c^2} \,c \Int dt' \, \frac{... ...)}\,\deL{R(\vx,t')} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right)\right\}$$

$$=- \frac{q}{4\pi \vepsilon_0} \Int dt' \, \frac{\bm{\beta}(t')}{R... ...)}\,\deL{R(\vx,t')} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right)\right\}$$

$$=-\frac{q}{4\pi \vepsilon_0}\frac{1}{c} \Int dy \,\di{t'}{y} \fra... ...} \deL{y}\delta(t-y),\qquad y= t' + \frac{\left\vert\vx-\vx'(t')\right\vert}{c}$$

$$=\frac{q}{4\pi \vepsilon_0}\frac{1}{c} \Int dy \,\deL{y}\left(\di... ...c{\bm{\beta}(t')}{R(\vx,t')}\right)\delta(t-y) \qquad \because\,\hbox{部分積分}$$

$$=\frac{q}{4\pi \vepsilon_0}\frac{1}{c}\deL{t}\left[\di{t'}{t} \fr... ...deL{t'}\left[\di{t'}{t} \frac{\bm{\beta}(t')}{R(\vx,t')}\right]_{h(\vx,y,t')=0}$$

$$=\frac{q}{4\pi \vepsilon_0}\frac{1}{c}\frac{1}{\kappa(t')}\deL{t'... ...ft[\frac{1}{\kappa(t')} \frac{\bm{\beta}(t')}{R(\vx,t')}\right]_{h(\vx,t,t')=0}$$

を得る。