推迟势电场速度项：
\(\mathbf{E}_{\text{vel}}(\mathbf{r}, t) = \frac{q}{4\pi\epsilon_0} \left[ \frac{(\hat{n} - \bm{\beta})(1 - \beta^2)}{(1 - \hat{n} \cdot \bm{\beta})^3 R^2} \right]_{t_r}\)
电荷沿 x 方向匀速运动，速度为 \(v\)，则：
\(\mathbf{E} = \frac{q}{4\pi\epsilon_0} \cdot \frac{ \left( \gamma (x - vt),\ y,\ z \right) }{ \left[ \gamma^2 (x - vt)^2 + y^2 + z^2 \right]^{3/2} }\)
\(\beta = v/c\),
\(R = \gamma^2(x - vt)^2 + y^2 + z^2\)
\(\nabla^2 E\)的分量：
\[ \nabla^2 \mathbf{E} = \frac{3q \gamma^2 v^2}{4\pi\epsilon_0 c^2 R^{7/2}} \begin{pmatrix} \gamma (x-vt) \left[ 2\gamma^2 (x-vt)^2 - 3(y^2 + z^2) \right] \\ y \left[ 4\gamma^2 (x-vt)^2 - y^2 - z^2 \right] \\ z \left[ 4\gamma^2 (x-vt)^2 - y^2 - z^2 \right] \end{pmatrix} \]
而:
\[ \boxed{ \frac{\partial^2 \mathbf{E}}{\partial t^2} = \frac{3q \gamma^2 v^2}{4\pi\epsilon_0 R^{7/2}} \cdot \begin{pmatrix} \gamma (x - vt) \left( 2\gamma^2 (x - vt)^2 - 3(y^2 + z^2) \right) \\ y \left( 4\gamma^2 (x - vt)^2 - y^2 - z^2 \right) \\ z \left( 4\gamma^2 (x - vt)^2 - y^2 - z^2 \right) \end{pmatrix} } \]
可见:
\(\frac{\partial^2 E}{\partial t^2} =c^2 \nabla^2 E\)
推迟势磁场速度项:
\(\mathbf{B}_{\text{vel}} = \frac{q}{4\pi\epsilon_0 c} \left[ \frac{c(\hat{n} \times \bm{\beta})}{(1 - \hat{n} \cdot \bm{\beta})^2 R^2} \right]_{t_r}\)
匀速时：
\(\mathbf{B} = \frac{q(1 - \beta^2)}{4\pi\epsilon_0 c^2} \frac{\mathbf{v} \times \mathbf{R}}{D^{3/2}}\)
\(\mathbf{R} = (x - vt, y, z)\)
分量形式：
\(B_x = 0 \)
\(B_y = -\frac{q(1 - \beta^2)v}{4\pi\epsilon_0 c^2} \frac{z}{D^{3/2}} \)
\(B_z = \frac{q(1 - \beta^2)v}{4\pi\epsilon_0 c^2} \frac{y}{D^{3/2}} \)
\[ \nabla^2 \mathbf{B} = \frac{3\mu_0 q \gamma^2 v^3}{4\pi c^2 R^{7/2}} \begin{pmatrix} 0 \\ - z (4\gamma^2 u^2 - y^2 - z^2) \\ y (4\gamma^2 u^2 - y^2 - z^2) \end{pmatrix} \]
也满足：
\(\frac{\partial^2 \mathbf{B}_{\text{vel}}}{\partial t^2} = c^2 \nabla^2 \mathbf{B}_{\text{vel}}\)