3 ベクトルポテンシャルの時間微分

$ \vA \xt$$ t$ で偏微分すると、

$\displaystyle \deL{t} \vA\xt$ $\displaystyle = \frac{q\mu_0}{4\pi} \Int dt'   \deL{t} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right) \frac{\vu(t')}{R(\vx,t')}\right\}$    
  $\displaystyle = -\frac{q}{4\pi \vepsilon_0}\frac{1}{c^2}  c \Int dt'   \frac{...
...)} \deL{R(\vx,t')} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right)\right\}$    
  $\displaystyle =- \frac{q}{4\pi \vepsilon_0} \Int dt'   \frac{\bm{\beta}(t')}{R...
...)} \deL{R(\vx,t')} \left\{ \delta\left(t'-t+\frac{R(\vx,t')}{c}\right)\right\}$    
  $\displaystyle =-\frac{q}{4\pi \vepsilon_0}\frac{1}{c} \Int dy  \di{t'}{y} \fra...
...} \deL{y}\delta(t-y),\qquad y= t' + \frac{\left\vert\vx-\vx'(t')\right\vert}{c}$    
  $\displaystyle =\frac{q}{4\pi \vepsilon_0}\frac{1}{c} \Int dy  \deL{y}\left(\di...
...c{\bm{\beta}(t')}{R(\vx,t')}\right)\delta(t-y) \qquad \because \hbox{部分積分}$    
  $\displaystyle =\frac{q}{4\pi \vepsilon_0}\frac{1}{c}\deL{t}\left[\di{t'}{t} \fr...
...deL{t'}\left[\di{t'}{t} \frac{\bm{\beta}(t')}{R(\vx,t')}\right]_{h(\vx,y,t')=0}$    
  $\displaystyle =\frac{q}{4\pi \vepsilon_0}\frac{1}{c}\frac{1}{\kappa(t')}\deL{t'...
...ft[\frac{1}{\kappa(t')} \frac{\bm{\beta}(t')}{R(\vx,t')}\right]_{h(\vx,t,t')=0}$    

を得る。

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp