# Maxwell's Equations, cast in terms of magnetic vector potential [closed]

Derive $$\nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \sigma \frac{\partial A}{\partial t} + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = 0$$ where $D_r$ is a "remnant displacement field" similar to the material polarization, $$D = \epsilon_0 \epsilon_r E + D_r$$

A numerical package (COMSOL Multiphysics 4.3b) solves maxwell's equations in the time domain in its transient electromagnetic wave module using this equation. The "remnant displacement field" is used to model the nonlinear polarization of a material.

My attempt...

$$\nabla \times H = J_f + \frac{\partial D}{\partial t}$$ $$\nabla \times \frac{1}{\mu_0 \mu_r} \nabla \times A = J_f + \epsilon_0 \epsilon_r \frac{\partial E}{\partial t} + \frac{\partial D_r}{\partial t}$$ $$\nabla \times \frac{1}{\mu_r} \nabla \times A = \mu_0 J_f + \mu_0 \frac{\partial D_r}{\partial t} + \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t}\left( -\nabla V - \frac{\partial A}{\partial t} \right)$$ $$\nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = \mu_0 \left( J_f - \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t} \nabla V \right)$$ $$J_f = \sigma E = \sigma \left(-\nabla V - \frac{\partial A}{\partial t} \right)$$ Substitute $J_f$... $$\nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \sigma \frac{\partial A}{\partial t} + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = \mu_0 \left( -\sigma \nabla V - \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t} \nabla V \right)$$ Any ideas on why the right side goes to zero?

## closed as off-topic by ACuriousMind♦, Kyle Kanos, BMS, JamalS, Pranav HosangadiJan 8 '15 at 8:45

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• Can you provide the source from where you are getting that equation? – glS Jan 7 '15 at 22:04
• $\nabla V$ is a conservative force. Could it be that you have no such thing in your case? – Sofia Jan 7 '15 at 22:38
• The equation is, unfortunately, not clearly spelled out in their online documentation. It is only available in the software. I am not sure if $\nabla V$ is zero. This is a purely electrodynamic problem---that is, there are no static charges. This problem is an electric field propagating through a nonlinear medium (the wave equation is nonlinear with a source term proportional to $E^2$). – kordon Jan 8 '15 at 1:12

Let $\lambda$ be defined such that, $$\frac{\partial}{\partial t} \nabla \lambda = \nabla V$$ then, $$V^\prime = V - \frac{\partial \lambda}{\partial t}$$ $$\nabla V^\prime = \nabla V - \frac{\partial}{\partial t} \nabla \lambda = \nabla V - \nabla V = 0$$ You must also transform the magnetic vector potential. $$A^\prime = A + \nabla \lambda$$
However, as can be read in this link (page 577), electrodynamic problems in this software implicitly use this gauge. The $A$ used in the question's equation is untransformed; however, it is transformed in the software when implementing the routine.