Boundary conditions for displacement vector D

Griffith's writes in chapter 7 electrodynamics that D1.a - D2.a = sigma. a.
But minus sine comes when we evaluate the dot product first.
How does the minus sign occur without evaluating the dot product?

vanhees71
Gold Member
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You have simply to do the integral over the Gaussian "pill box" shown in Fig. 7.48 in Griffiths's book (4th edition). Making the pillbox very small, so that ##\vec{D}## on both sides of the boundary surface can be taken as constant along the area ##a##. The contributions from the four surfaces of the pill box perpendicular to the boundary surface cancel pairwise, but there may be a surface charge along the boundary surface, and then, even if you make the height of the pill box arbitrarily small, you always get a non-zero result, namely the total charge within the pill box, which is ##\sigma a##, and thus you have
$$\vec{n} \cdot (\vec{D}_1-\vec{D}_2)=\sigma.$$
Here ##\vec{D}_1## (##\vec{D}_2##) denotes the value of ##\vec{D}## when approaching the point on the boundary surface under investigation from side 1 (side 2) of the boundary surface. The minus in the above equation comes from the fact that the surface normal vector of the pill box parallel to the boundary at side 2 is just ##-\vec{n}##, where ##\vec{n}## is the direction of the surface normal vector at side 1 (see again Fig. 7.48, where in my notation ##\vec{a}=a \vec{n}##).

Still confusing!

vanhees71
$$a \vec{n} \cdot \vec{D}_1 + a (-\vec{n}) \cdot \vec{D}_2 = a \vec{n} \cdot (\vec{D}_1-\vec{D}_2).$$