Interface conditions for electromagnetic fields

Maxwell's equations describe the behaviour of electromagnetic fields; electric field, electric displacement field, and the magnetic field. The differential forms of these equations require that there is always an open neighbourhood around the point to which they are applied, otherwise the vector fields E, D, B and H are not differentiable. In other words, the medium must be continuous. On the interface of two different media with different values for electrical permittivity and magnetic permeability, that condition does not apply.

However the interface conditions for the electromagnetic field vectors can be derived from the integral forms of Maxwell's equations.

Interface conditions for electric field vectors

For electric field

\mathbf {n} _{12}\times (\mathbf {E} _{2}-\mathbf {E} _{1})=\mathbf {0}

where:
$\mathbf {n} _{12}$ is normal vector from medium 1 to medium 2.

Therefore, the tangential component of E is continuous across the interface.

Outline of proof from Faraday's law

We begin with the integral form of Faraday's law:

\oint _{\partial \Sigma }\mathbf {E} \cdot d{\boldsymbol {\ell }}=-\int _{\Sigma }{\frac {\partial \mathbf {B} }{\partial t}}\cdot d\mathbf {A}

Choose

\Sigma

as a small square across the interface. Then, have the sides perpendicular to the interface shrink to infinitesimal length. The area of integration now looks like a line, which has zero area. In other words:

\lim _{linelength\to 0}\mathbf {A} =0

Since

\partial \mathbf {B} /\partial t

remains finite in this limit, the whole right hand side goes to zero. All that is left is:

\oint _{\partial \Sigma }\mathbf {E} \cdot d{\boldsymbol {\ell }}=0

Two of our sides are infinitesimally small, leaving only

\int _{medium1}\mathbf {E} \cdot d{\boldsymbol {\ell }}+\int _{medium2}\mathbf {E} \cdot d{\boldsymbol {\ell }}=0

Assuming we made our square small enough that E is roughly constant, its magnitude can be pulled out of the integral. As the remaining sides to our original loop, the

d{\boldsymbol {\ell }}

in each region run in opposite directions, so we define one of them as the tangent unit vector

{\boldsymbol {t}}

and the other as

-{\boldsymbol {t}}

(\mathbf {E} _{2}\cdot {\boldsymbol {t}})l-(\mathbf {E} _{1}\cdot {\boldsymbol {t}})l=\mathbf {0}

After dividing by l, are rearranging,

(\mathbf {E} _{2}-\mathbf {E} _{1})\cdot {\boldsymbol {t}}=\mathbf {0}

This argument works for any tangential direction. The difference in electric field dotted into any tangential vector is zero, meaning only the components of $\mathbf {E}$ parallel to the normal vector can change between mediums. Thus, the difference in electric field vector is parallel to then normal vector. Two parallel vectors always have a cross product of zero.

\mathbf {n} _{12}\times (\mathbf {E} _{2}-\mathbf {E} _{1})=\mathbf {0}

For electric displacement field

(\mathbf {D} _{2}-\mathbf {D} _{1})\cdot \mathbf {n} _{12}=\rho _{s}

where:
$\mathbf {n} _{12}$ is normal vector from medium 1 to medium 2.
$\rho _{s}$ is the surface charge between the media (unbounded charges only, not coming from polarization of the materials).

Therefore, the normal component of D has a step of surface charge on the interface surface. If there is no surface charge on the interface, the normal component of D is continuous.

Interface conditions for magnetic field vectors

For magnetic field

(\mathbf {B} _{2}-\mathbf {B} _{1})\cdot \mathbf {n} _{12}=0

where:
$\mathbf {n} _{12}$ is normal vector from medium 1 to medium 2.

Therefore, the normal component of B is continuous across the interface.

For magnetic field strength

\mathbf {n} _{12}\times (\mathbf {H} _{2}-\mathbf {H} _{1})=\mathbf {j} _{s}

where:
$\mathbf {n} _{12}$ is normal vector from medium 1 to medium 2.
$\mathbf {j} _{s}$ is the surface current density between the two media (unbounded current only, not coming from polarisation of the materials).

Therefore, the tangential component of H is continuous across the surface if there's no surface current present.