速度项：
\[ \mathbf{E}_{\text{vel}}(\mathbf{r}, t) = \frac{q}{4\pi\epsilon_0} \left[ \frac{(\hat{n} - \bm{\beta})(1 - \beta^2)}{(1 - \hat{n} \cdot \bm{\beta})^3 R^2} \right]_{t’} \]
辐射项：
\[ \mathbf{E}_{\text{rad}} = \frac{q}{4\pi\epsilon_0} \frac{\hat{n} \times \left[ (\hat{n} - \bm{\beta}) \times \dot{\bm{\beta}} \right]}{c(1 - \hat{n}\cdot\bm{\beta})^3 R} \] 在沿x方向运动时速度项分量：
\[ E_x = \frac{q(1 - \beta^2)}{4\pi\epsilon_0} \frac{\frac{x - x_q(t’)}{R} - \beta}{\left[ R (1 - \beta \frac{x - x_q(t’)}{R}) \right]^3} \] \[ E_y = \frac{q(1 - \beta^2)}{4\pi\epsilon_0} \frac{y/R}{\left[ R (1 - \beta \frac{x - x_q(t’)}{R}) \right]^3} \] \[ E_z = \frac{q(1 - \beta^2)}{4\pi\epsilon_0} \frac{z/R}{\left[ R (1 - \beta \frac{x - x_q(t’)}{R}) \right]^3} \]
令 \( \gamma = (1 - \beta^2)^{-1/2} \)，并定义： \[ \mathcal{D} = R \left( 1 - \beta \frac{x - x_q(t’)}{R} \right) = R - \beta (x - x_q(t’)) \] 则电场分量可写为： \[ E_{x,vel} = \frac{q}{4\pi\epsilon_0} \frac{(x - x_q(t’)) - \beta R}{\gamma^2 \mathcal{D}^3} \] \[ E_{y,vel} = \frac{q}{4\pi\epsilon_0} \frac{y}{\gamma^2 \mathcal{D}^3} \] \[ E_{z,vel} = \frac{q}{4\pi\epsilon_0} \frac{z}{\gamma^2 \mathcal{D}^3} \]
辐射场分量： \[ E_x = -\frac{q}{4\pi\epsilon_0} \frac{\dot{\beta}(y^2 + z^2)}{cR^3(1 - \beta\frac{x - x_q}{R})^3} \] \[ E_y = \frac{q}{4\pi\epsilon_0} \frac{\dot{\beta}(x - x_q)y}{cR^3(1 - \beta\frac{x - x_q}{R})^3} \] \[ E_z = \frac{q}{4\pi\epsilon_0} \frac{\dot{\beta}(x - x_q)z}{cR^3(1 - \beta\frac{x - x_q}{R})^3} \]
或：
\[ \boxed{ \begin{aligned} E_{x,rad} &= -\frac{q a}{4\pi\epsilon_0 c^2} \cdot \frac{y^2 + z^2}{\mathcal{D}^3} \\ E_{y,rad} &= \frac{q a(t_r)}{4\pi\epsilon_0 c^2} \cdot \frac{(x - x_q) y}{\mathcal{D}^3} \\ E_{z,rad} &= \frac{q a(t_r)}{4\pi\epsilon_0 c^2} \cdot \frac{(x - x_q) z}{\mathcal{D}^3} \end{aligned} } \]
有：
\(E_{x,rad} = -sss\)
\(E_{y,rad} = \frac{a}{c^2} \cdot (x - x_q) E_{y,vel} \)
\(E_{z,rad} = \frac{a}{c^2} \cdot (x - x_q) E_{z,vel} \)
\(\dot{\beta}=\frac{a}{c}\)
\(x_q = x_q(t_r) = \frac{c^2}{\alpha} \left( \sqrt{1 + \left( \frac{\alpha t_r}{c} \right)^2} - 1 \right)\)
\(t_r = t - \frac{1}{c} \sqrt{ (x - x_q(t_r))^2 + y^2 + z^2 }\)
\(\alpha\)为固有加速度，也就是电荷感受到的作用力转换的加速度
使用\(x=\frac{1}{2}at^2\)来计算，是非相对论极限下的值
实验室中的加速度\(a=\frac{dv}{dt} = \alpha \left( 1 - \frac{v^2}{c^2} \right)^{3/2}\)
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$$ \mathbf{E} = \frac{q}{4\pi \epsilon_0} \left( \frac{\hat{R} - \boldsymbol{\beta}}{\gamma^2 (1 - \hat{R} \cdot \boldsymbol{\beta})^3 R^2} + \frac{\hat{R} \times \left( (\hat{R} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}} \right)}{c (1 - \hat{R} \cdot \boldsymbol{\beta})^3 R} \right)_{\text{ret}} $$
沿 x 轴运动:
$$ \mathbf{E}(\mathbf{r}, t) = \frac{q}{4\pi \epsilon_0} \left[ \frac{\hat{R} - \frac{v_q}{c} \hat{x}}{\gamma^2 \left(1 - \frac{v_q}{c} \frac{x - x_q}{R}\right)^3 R^2} + \frac{1}{c} \frac{ \frac{a_q}{c^2} }{ \left(1 - \frac{v_q}{c} \frac{x - x_q}{R}\right)^3 R } \left( \frac{x - x_q}{R} \hat{R} - \hat{x} \right) \right]_{\text{ret}} $$
$$ \mathbf{E}(\mathbf{r}, t) = \frac{q}{4\pi \epsilon_0} \left[ \frac{\hat{R} - \frac{\mathbf{v}_q}{c}}{\gamma^2 \left(1 - \frac{\mathbf{v}_q \cdot \hat{R}}{c}\right)^3 R^2} + \frac{\hat{R} \times \left( (\hat{R} - \frac{\mathbf{v}_q}{c}) \times \frac{\dot{\mathbf{v}}_q}{c} \right)}{c \left(1 - \frac{\mathbf{v}_q \cdot \hat{R}}{c}\right)^3 R} \right]_{\text{ret}} $$
\(v_q(t’)=\dot{x_q}(t’)=\frac{d x_q(t’)}{dt’}\),
\(a_q(t’)=\dot{v_q}(t’)=\frac{d v_q(t’)}{dt’}\)