# Boundary conditions for Maxwell's equations at the interface between two media

Consider the following simple Maxwell's equations: $$\nabla\cdot\mathrm{D}=\rho$$ $$\nabla\times\mathrm{E}+i\omega\mathrm{B}=0$$ $$\nabla\cdot\mathrm{B}=0$$ $$\nabla\times\mathrm{H}=\mathrm{J}+i\omega\mathrm{D}$$

It is well known that at the interface between two media, the boundary condition gives $$n\times(\mathrm{E}_1-\mathrm{E}_2)=0$$ $$n\cdot(\mathrm{D}_1-\mathrm{D}_2)=\sigma$$ $$n\times(\mathrm{H}_1-\mathrm{H}_2)=\mathrm{J}$$ $$n\cdot(\mathrm{B}_1-\mathrm{B}_2)=0$$

Now my question is: are those boundary conditions independent or redundant? Could someone explain them to me? Thank you!

Each boundary condition comes from an independent Maxwell equation, so the four boundary conditions are independent. The right-hand side of the third boundary condition should K, the surface current. The first boundary condition could be replaced by $\phi_1=\phi_2$, which is easier to implement
This question is natural, and I needed to solve it for myself many years ago. I don't remember the derivation and don't have time to reproduce it, but at least in the case of $J=\rho=\sigma=0$ and fixed $\omega\neq 0$ the boundary conditions are not independent, and it is enough to require that tangential components of $E$ and $H$ are continuous at the boundary. I suspect this result is well-known, but I don't have a reference.