# Continuity conditions in electrodynamics.

1. Mar 10, 2013

### nick4189

I have a question about the derivation of the boundary conditions at surfaces of electromagnetic fields. These conditions say, that the tangential component of the electric and the normal component of the magnetic field are continuous at surfaces.
Their derivation goes as follows: To derive them for the electric field,
one starts with the Maxwell equation $\mbox{rot } E=- \frac{1}{c} \frac{\partial B}{\partial t}$ and uses Stokes theorem for a line integral of $E$ across the boundary, as it is depicted at http://ocw.mit.edu/courses/electric...applications-fall-2005/lecture-notes/lec2.pdf. The derivation for the magnetic field goes similar, but now we have to use the equation $\mbox{div} B=0$.
My question now is: Why can't we use some other Maxwell equations as well, to obtain further conditions? We still have the Maxwell equation $\mbox{rot} B= \frac{4 \pi}{c}j+ \frac{1}{c} \frac{\partial E}{\partial t}$. Shouldn't we get something out of it at least for $j=0$? What is the reason that there isn't something similar, that works for other Maxwell equations?

2. Mar 10, 2013

### jasonRF

It turns out that using similar arguments for the other two Maxwell equations is indeed an important part of figuring out boundary condition - you are on the right track! What may not be obvious, is that for each of the two remaining equations we usually allow for the existence of a surface source.

More specifically, for $\mathbf{\nabla\cdot D} = \rho$ we integrate over a small "pill-box", just like you did for the continuity of the perpendicular component of B. But we allow for the existence of a surface charge density $\rho_s$ (units of charge per area), so that the condition for the perpendicular component of D across the boundary should read, $\mathbf{\hat{n}\cdot}\left(\mathbf{D}_2-\mathbf{D}_1 \right)=\rho_s$ where $\mathbf{\hat{n}}$ is a unit vector pointing outward from medium 1. You should make sure you can derive this - I used MKS so you can derive it in CGS. Anyway, physically this simply means that a surface charge density will cause a discontinuity in the perpendicular component of D. Note that for a perfect conductur, the electric field inside vanishes, and hence the surface charge density on the surface of the conductor can be related to the normal component of the exterior electric field; if it is free space then $\mathbf{D}=\epsilon_0 \mathbf{E}$, so the surface charge density is simply $\epsilon_0 E_n$ where the field is evaluated on the surface of the conductor.

Likewise for the $\mathbf{\nabla \times H}= \mathbf{J}+ \partial_t \mathbf{D}$, we assume a surface current density $\mathbf{J}_s$ (units are current per length) and we get $\mathbf{\hat{n}\times}(\mathbf{H}_2 - \mathbf{H}_1) = \mathbf{J}_s$. Again, I used MKS and you should derive this yourself with your favorite units. PHysically, this means that a surface current will create a disconinuity in the parallel component of H.

I hope that helps!

jason