- #1
Maniac_XOX
- 86
- 5
- TL;DR
- Is it possible to derive equations for electric field E and magnetic B from the following equation?
$$\Box A_\alpha +\mu^2 A_\alpha = 2\beta A_\mu \partial_\alpha A^\mu + \frac {4\pi}{c} J_\alpha$$
where ##A=(\Phi, \vec A)## and ##J=(\rho, \vec J)##
using a static configuration first where ##α=0##
and then a dynamic one where ##α=i##
knowing that ##E= - \nabla^2 \Phi - \frac {\partial A}{dt}## and ##B= \nabla \times A## and ##-\nabla^2 A + \nabla (\nabla A) = \nabla \times (\nabla \times A)##
I personally tried but because of the ##2\beta A_\mu## term i cannot connect these
My attempts so far:
For static configuration $$- \frac {\partial \Phi}{c^2 \partial t^2} + \nabla^2 \Phi +\mu^2 \Phi = 2\beta A_\mu \frac {\partial A^\mu}{\partial t} + \frac {4\pi}{c} \rho$$
For dynamic configuration $$- \frac {\partial \vec A}{c^2 \partial t^2} + \nabla^2 \vec A +\mu^2 \vec A = 2\beta A_\mu \nabla \vec A + \frac {4\pi}{c} \vec J$$
where ##A=(\Phi, \vec A)## and ##J=(\rho, \vec J)##
using a static configuration first where ##α=0##
and then a dynamic one where ##α=i##
knowing that ##E= - \nabla^2 \Phi - \frac {\partial A}{dt}## and ##B= \nabla \times A## and ##-\nabla^2 A + \nabla (\nabla A) = \nabla \times (\nabla \times A)##
I personally tried but because of the ##2\beta A_\mu## term i cannot connect these
My attempts so far:
For static configuration $$- \frac {\partial \Phi}{c^2 \partial t^2} + \nabla^2 \Phi +\mu^2 \Phi = 2\beta A_\mu \frac {\partial A^\mu}{\partial t} + \frac {4\pi}{c} \rho$$
For dynamic configuration $$- \frac {\partial \vec A}{c^2 \partial t^2} + \nabla^2 \vec A +\mu^2 \vec A = 2\beta A_\mu \nabla \vec A + \frac {4\pi}{c} \vec J$$
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