Квантовая нелинейная электродинамика

Last modified by Alexey Popov on 2019/12/17 15:38

$\mathbf E = -\nabla \varphi -\frac{\partial \mathbf A}{\partial t} - \nabla \times \mathbf C - i\frac{\hbar}{m_e}\left (\Delta \mathbf A - \frac{1}{c^2}\frac{\partial^2 \mathbf A}{\partial t^2}\right ) - i\left (\Delta \mathbf E_q \pm \nabla \times \nabla \times \mathbf E_q - \frac{1}{v^2}\frac{\partial^2 \mathbf E_q}{\partial t^2}\right )$

$\mathbf B = \nabla \times \mathbf A - \frac{1}{c^2}\frac{\partial \mathbf C}{\partial t} - \frac{1}{c^2}\nabla \xi - i\frac{1}{c^2}\frac{\hbar}{m_e}\left (\Delta \mathbf C - \frac{1}{c^2}\frac{\partial^2 \mathbf C}{\partial t^2}\right ) - i\left (\Delta \mathbf B_q \pm \nabla \times \nabla \times \mathbf B_q - \frac{1}{v^2}\frac{\partial^2 \mathbf B_q}{\partial t^2}\right )$

$\nabla \cdot \mathbf A + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = -i\frac{1}{c^2}\frac{\hbar}{m_e}\left (\Delta\varphi - \frac{1}{c^2}\frac{\partial^2 \varphi}{\partial t^2}\right )$

$\nabla \cdot \mathbf C + \frac{1}{c^2}\frac{\partial \xi}{\partial t} = -i\frac{1}{c^2}\frac{\hbar}{m_e}\left (\Delta\xi - \frac{1}{c^2}\frac{\partial^2 \xi}{\partial t^2}\right )$

$\nabla \cdot \mathbf E_q = \varphi$

$\nabla \cdot \mathbf B_q = \frac{1}{v^2}\xi$

$\nabla \times \mathbf E_q = - \frac{\partial \mathbf B_q}{\partial t}$

$\nabla \times \mathbf B_q = \mathbf A + \frac{1}{v^2}\frac{\partial \mathbf E_q}{\partial t}$

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Created by Alexey Popov on 2019/12/16 13:51