推迟势总结与洛伦兹变换
推迟势中的坐标系关系(下图rR’得x位置是错误的): 其中: \( R = \sqrt{(x - vt’)^2 + y^2 + z^2} \) \( R’ = \sqrt{(x - vt)^2 + \frac{y^2 + z^2}{\gamma^2}} \), 或:\( \gamma R’ = \sqrt{(x - vt)^2 \gamma^2 + y^2 + z^2} \), 其中 \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) 求解过程如下: 电荷q在位置\( r’(vt’,0,0,t’) \)处发出光子,到达点\( r(x,y,z,t) \), 此时满足关系: \( c(t-t’)=R=\sqrt{(x - vt’)^2 + y^2 + z^2} \) 于是可以得到t和t’的关系: \[ t=t’+\sqrt{(x - vt’)^2 + y^2 + z^2}/c \\ t’ = \frac{c^2 t - x v - c\sqrt{(x - v t)^2 + (1 - v^2/c^2)(y^2 + z^2)}}{c^2 - v^2} \] ...