# How do you show that the normal derivative of the magnetic vector potential is discontinous across a surface current?

This is from Griffith's "Introduction to Electrodynamics" 4th edition, problem 5.33.

I need to show that

$$\tag 1 \frac{\partial \textbf{A}_{above}}{\partial n}-\frac{\partial \textbf{A}_{below}}{\partial n}=-\mu_{0}\textbf{K}$$

where $\textbf{A}$ is the magnetic vector potential (i.e. $\textbf{B}=\nabla \times \textbf{A}$), and $\textbf{K}$ is the surface current density, and I am supposed to use the equations:

$$\tag 2 \textbf{B}_{above}-\textbf{B}_{below}=\mu_{0}(\textbf{K}\times\hat{\textbf{n}})$$

$$\tag 3 \textbf{A}_{above}=\textbf{A}_{below}$$

$$\tag 4 \nabla\cdot\textbf{A}=0$$

The problem is, I'm not sure what $\frac{\partial\textbf{A}}{\partial n}$ is. For a scalar, the book defines $\frac{\partial V}{\partial n}$ as: $\nabla V \cdot \hat{\textbf{n}}$

Would $\frac{\partial\textbf{A}}{\partial n}$ simply a vector with components $\nabla A_{i}\cdot\hat{\textbf{n}}$ for $i=x,y,z$?

That's what I tried, but with no luck so far.

Multiplying vectorially with $\mathbf n$ your (2), and using the identity $$\tag A \mathbf A \times ( \mathbf B \times \mathbf C) = (\mathbf A \cdot \mathbf C) \mathbf B - (\mathbf A \cdot \mathbf B) \mathbf C$$ you obtain $$\tag B (\mathbf B_{\text{above}} - \mathbf B_{\text{below}})\times \mathbf n = - \mu_0 \mathbf K.$$
Now use in both the terms in (B) the identity $$\tag C ( \nabla \times \mathbf A) \times \mathbf n = (\mathbf n \cdot \nabla)\mathbf A - \nabla (\mathbf A \cdot \mathbf n),$$
and remember that the normal derivative is defined (or equivalently, can be expressed) as $$\frac{\partial \mathbf A}{\partial n} \equiv (\mathbf n \cdot \nabla) \mathbf A.$$
• How is this term $\nabla (\mathbf A \cdot \mathbf n)$ gonna cancel? Commented May 8 at 7:31