Magnetic Vector Potential prob 5.32 Griffiths

[precise]

Please refer to the problem towards the end of page in the following link. It's related to discontinuity in normal derivative of magnetic vector potential across a current carrying surface. Prob 5.32 in Griffths.

http://physicspages.com/2013/04/08/magnetostatic-boundary-conditions/

The solution says "Since A is continuous across the surface, the derivatives in directions perpendicular to the surface (that is, {x} and {y}) will be the same on both sides."

I couldn't understand the above argument which says the derivatives along x and y are same across the boundary. What is the calculus I'm missing. Could you explain in detail, or suggest some reading for that math part I'm missing.

I found this thread, with same concern. It's still not clear why though.

https://www.physicsforums.com/showthread.php?t=498682

Thanks

 

[vanhees71]

While your weblink is a bit handwaving to my taste, it's very clearly written in Griffiths's book. The magnetic field has only a finite jump, not a more severe singularity. From
$$\vec{B}=\vec{\nabla} \times \vec{A}$$
you get for any surface $F$ with boundary $\partial F$.
$$\int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}=\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{A}.$$
By the usual arguments with an infinitesimal rectangle penetrating the surface perpendicularly and from the finiteness of $\vec{B}$ along this rectangle you get that in the limit of a vanishing height the surface integral is 0 and so also the line integral over the vector potential. This implies that the tangential components of the vector potential must be continuous.

From the Coulomb-gauge constraint
$$\vec{\nabla} \cdot \vec{A}=0$$
it follows in a similar way by using Gauß's integral theorem that also the normal components of the vector potential are continuous across the surface.

The tangential components of the magnetic field, however must have a jump if a surface current is present as in Griffiths's example. This implies that the derivatives of $\vec{A}$ must have these jumps since $\vec{B}=\vec{\nabla} \times \vec{A}$.

Now I'd go along as follows. Write down Ampere's Law for the above described infinitesimal square, which you orient such that its surface-normal vector is in direction of the surface current $\vec{K}$. Then you get
$$\mu I=\int_F \mathrm{d}^2 \vec{x} (\vec{\nabla} \times \vec{B})=-\int_F \mathrm{d}^2 \vec{x} \cdot \vec{\nabla}^2 \vec{A}.$$
This will immediately give
$$(\vec{n} \cdot \vec{\nabla}) \vec{A}=-\mu_0 \vec{K}.$$
You just have to work out the last integral above in the limit of an infinitesimal square!