В поисках магнитного монополя

Last modified by Alexey Popov on 2019/07/01 18:17

$\mathbf A = \frac{1}{c^2}\varphi \mathbf v$

$\nabla \cdot \mathbf A = - \frac{1}{c^2}\frac{\partial \varphi}{\partial t}$

$\mathbf E = - \nabla \varphi - \frac{\partial \mathbf A}{\partial t}$

$\mathbf B = \nabla \times \mathbf A = \nabla \times \varphi \mathbf v = \frac{1}{c^2}\varphi (\nabla \times \mathbf v) + \frac{1}{c^2}(\nabla \varphi \times \mathbf v)$

$\nabla \cdot \mathbf B = \frac{1}{c^2}\nabla \cdot (\varphi \boldsymbol \omega) + \frac{1}{c^2}\nabla \cdot ((\nabla \varphi \times \mathbf v)) = \frac{1}{c^2}\varphi \nabla \cdot \boldsymbol \omega + \frac{1}{c^2} \nabla\varphi \cdot \boldsymbol \omega + \frac{1}{c^2}(\nabla \times \nabla \varphi) \cdot \mathbf v - \frac{1}{c^2}(\nabla \varphi\cdot \boldsymbol \omega) = \frac{1}{c^2} \nabla\varphi \cdot \boldsymbol \omega - \frac{1}{c^2}(\nabla \varphi\cdot \boldsymbol \omega) = 0$

$\mathbf B = \nabla \times \mathbf A - \frac{1}{c^2}\varphi \boldsymbol \omega$

$\nabla \cdot \mathbf B = \frac{1}{c^2}(\nabla \varphi\cdot \boldsymbol \omega) = -\frac{1}{c^2}(\mathbf E \cdot \boldsymbol \omega) + \frac{1}{c^2}(\frac{\partial \mathbf A}{\partial t}\cdot \boldsymbol \omega) =\\= -\frac{1}{c^2}\mathbf E \cdot \boldsymbol \omega + \frac{1}{c^2}\frac{\partial \mathbf A\cdot \boldsymbol \omega}{\partial t} - \frac{1}{c^2}\mathbf A\cdot\frac{\partial \boldsymbol \omega}{\partial t} = \rho_b$

$\rho_b\mathbf B = \mathbf f$

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Created by Alexey Popov on 2019/07/01 18:17