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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!

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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

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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.

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