Processing math: 100%

「４次元」

作成: 2017-11-16
更新: 2017-11-16

$\vec{A} \ =\ \left( \frac{\phi}{c} , \ A_x, \ A_y, \ A_z \right) \\ \vec{j} \ =\ \left( \rho c , \ i_x, \ i_y, \ i_z \right) \\ \left( \triangle - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \vec{A} \ =\ - \mu_0 \,\vec{j} \\$

$\square \ \equiv \ \triangle - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \\ \square \vec{A} \ =\ - \mu_0 \,\vec{j}$