题目
导出巡变电磁场的标势和矢势在洛伦兹规范条件下满足达朗贝尔方程:nabla^2 phi - (1)/(c^2) (partial^2 phi)/(partial t^2) = - (rho)/(varepsilon_0) ， nabla^2 vec(A) - (1)/(c^2) (partial^2 vec(A))/(partial t^2) = - mu_0 vec(J).
导出巡变电磁场的标势和矢势在洛伦兹规范条件下满足达朗贝尔方程:
$\nabla^2 \phi - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\varepsilon_0}$ ， $\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = - \mu_0 \vec{J}$.
题目解答
答案
根据麦克斯韦方程组及洛伦兹规范 $\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0$，可得：
1. 对高斯定律 $\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$，有：
\[
\nabla^2 \phi - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = -\frac{\rho}{\varepsilon_0}.
\]
2. 对安培-麦克斯韦定律 $\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}$，有：
\[
\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J}.
\]
最终结果为：
\[
\nabla^2 \phi - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = -\frac{\rho}{\varepsilon_0},
\]
\[
\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J}.
\]