Boundary conditions for $\mathbf D$ and $\mathbf H$

Question

I understand the derivation for the boundary conditions for $\mathbf B$ and $\mathbf E$ as it was explained to me in Griffiths, but Griffiths states the following:

$$H_{\text{above}}^{\bot} - H_{\text{below}}^{\bot} = -(M_{\text{above}}^{\bot}-M_{\text{below}}^{\bot})$$

and

$$\mathbf D_{\text{above}}^{\parallel} - \mathbf D_{\text{below}}^{\parallel} = \mathbf P_{\text{above}}^{\parallel} - \mathbf P_{\text{below}}^{\parallel}$$

This feels like the right idea, due to the relations that $\nabla \times \mathbf D = \nabla \times \mathbf P$ and $\nabla \cdot \mathbf H = - (\nabla \cdot \mathbf M)$, and from what I learned of the implications by $\nabla \cdot \mathbf E = \rho / \epsilon_0 $ implies a discontinuity in the parallel component of the electric field, which is why it makes sense to me that the discontinuity in the parallel component of the $\mathbf D$ field is the way it is. However, I can't quite piece together a coherent argument justifying this boundary conditions. How is this proven?

Also, the $\mathbf H$-field perpendicular component boundary conditions are expressed in magnitude in Griffiths, yet the $\mathbf D$-field parallel component boundary conditions were expressed as vectors, as I've shown. Why is that? If I made the $\mathbf H$-field a vector in the boundary condition I wrote for it, wouldn't that still be true?

Answer 1

In brief, the argument goes:

$\nabla \cdot {\bf D}$ equation $\rightarrow$ condition on $D_\perp$.

$\nabla \cdot {\bf B}$ equation $\rightarrow$ condition on $B_\perp$.

$\nabla \wedge {\bf E}$ equation $\rightarrow$ condition on $E_\parallel$.

$\nabla \wedge {\bf H}$ equation $\rightarrow$ condition on $H_\parallel$.

The first two arguments involve a cylinder-shaped Gaussian surface. The second two arguments involve an integral around a loop hugging the boundary. A complete argument will include the time-derivative terms and show that they are negligible in the limit where the Gaussian surface or the loop closely hugs the boundary. A complete argument will also include the surface conduction current in the $H_\parallel$ result, and the surface density of free charge in the $D_\perp$ result.

Once you have the above then you can get the equations for things like $E_\perp$, $D_\parallel$, $B_\parallel$ and $H_\perp$ simply by using the definitions of $\bf D$ and $\bf H$ in terms of ${\bf E}$, ${\bf P}$, ${\bf B}$, ${\bf M}$.

Answer 2

In response to the second part of your question, I think this is because there is only a single perpendicular component to the H and M fields, call it z, while the parallel fields must be specified by two components, call it x and y, because the boundary spans a plane.