Уравнения Максвелла с магнитными зарядами

Last modified by Alexey Popov on 2019/05/29 11:49

$\nabla \cdot \mathbf E = \frac{\rho_q}{\varepsilon_0} + \frac{Q_e}{m_e}\nabla \cdot (\mathbf A \times \mathbf B)+\frac{1}{c^2}\frac{Q_e}{m_e}\frac{\partial \mathbf A \cdot \mathbf E}{\partial t}+\mu_0\frac{Q_e}{m_e} \mathbf A \cdot \mathbf J\\$

$\nabla \cdot \mathbf B = \frac{1}{c^2}\frac{Q_e}{m_e}\nabla \cdot (\mathbf A \times \mathbf E)+\frac{1}{c^2}\frac{Q_e}{m_e}\frac{\partial \mathbf A \cdot \mathbf B}{\partial t}+\frac{1}{c^2}\frac{Q_e}{m_e} \mathbf E \cdot \mathbf B-\frac{1}{c^2}\frac{Q_e}{m_e} \mathbf A \cdot (\nabla \times \mathbf E)\\$

$\rho_m = \frac{1}{c^2}\frac{Q_e}{m_e} \mathbf E \cdot \mathbf B$

$\nabla \times \mathbf E = \frac{Q_e}{m_e}(\mathbf A \times (\nabla \times \mathbf B))- \frac{\partial \mathbf B}{\partial t} + \rho_m \mathbf v\\$

$\nabla \times \mathbf B = \frac{1}{c^2}\frac{Q_e}{m_e}(\mathbf A \times (\nabla \times \mathbf E)) +\frac{1}{c^2}\frac{Q_e}{m_e}(\mathbf E \times \mathbf B)+ \frac{1}{c^2}\frac{\partial \mathbf E}{\partial t}+\mu_0 \mathbf J\\$

$\nabla (\mathbf A \cdot \mathbf B) = - \frac{1}{c^2}\frac{\partial \mathbf A \times \mathbf E}{\partial t}$

$\nabla (\mathbf A \cdot \mathbf E) = - \frac{\partial \mathbf A \times \mathbf B}{\partial t}$

$\nabla (\nabla \cdot \mathbf B) = \frac{1}{c^2}\frac{Q_e}{m_e}\nabla(\nabla \cdot (\mathbf A \times \mathbf E))+\frac{1}{c^2}\frac{Q_e}{m_e}\frac{\partial \nabla(\mathbf A \cdot \mathbf B)}{\partial t}\\$

$\Delta \mathbf B +\frac{1}{c^2}\frac{Q_e}{m_e}\nabla \times (\mathbf A \times (\nabla \times \mathbf E)) +\frac{1}{c^2}\frac{Q_e}{m_e}\nabla \times (\mathbf E \times \mathbf B)+ \frac{1}{c^2}\frac{Q_e}{m_e}\frac{\partial (\mathbf A \times (\nabla \times \mathbf B))}{\partial t}+ \frac{1}{c^2}\frac{\partial^2 \mathbf B}{\partial t^2}+\mu_0 \nabla \times \mathbf J = \frac{1}{c^2}\frac{Q_e}{m_e} \Delta (\mathbf A \times \mathbf E)+ \frac{1}{c^2}\frac{Q_e}{m_e}\nabla \times (\nabla \times (\mathbf A \times \mathbf E))+\frac{1}{c^4}\frac{Q_e}{m_e}\frac{\partial^2 (\mathbf A \times \mathbf E)}{\partial t^2}\\$

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Created by Alexey Popov on 2019/02/22 10:36