1
$\begingroup$

I was reading in this pdf about the discontinuity of the derivative of the vector potential $\mathbf A$ across a boundary with $\hat{\mathbf n}$ acting as the normal to the plane having surface current density $\mathbf K$ and $\mathbf A_1$ & $\mathbf A_2$ being the value of vector potentials in medium $1$ and medium $2$ across the boundary.

[...] Finally let's put $\mathbf A$ in equation $(6)$ $$\boldsymbol \nabla \times (\mathbf A_1 -\mathbf A_2)~=~ \mu_0\mathbf K\times \hat{\mathbf n} $$ So, the derivatives of $\mathbf A$ have a discontinuity. But which one? Let's expand:

$$\hat{ \mathbf n}\times (\boldsymbol \nabla\times \mathbf A)~=~ n_i\boldsymbol \nabla A_i - \left(\hat{ \mathbf n}\cdot \boldsymbol \nabla \right)\mathbf A $$ Then $$n_i\boldsymbol\nabla \left(A_{i,1}-A_{i,2}\right)-\left(\hat{\mathbf n}\cdot \boldsymbol \nabla \right)(\mathbf A_1-\mathbf A_2) ~=~ \mu_0\mathbf K\tag{10}$$

But we have shown that each component of $\mathbf A$ is continuous at the surface. So the components of $$\boldsymbol\nabla \left(A_{i,1}-A_{i,2}\right)$$ parallel to the surface must be zero. Thus only the normal derivatives remain. Then the normal component of equation $(10)$ is identically zero, and the only non-zero components of the boundary condition are the tangential components $$\left(\hat{\mathbf n}\cdot \boldsymbol \nabla \right)(\mathbf A_1-\mathbf A_2)_\textrm{tan} ~=~ -\mu_0\mathbf K$$

...

I'm not comprehending the last reasoning they made above.

How does the continuity of $\mathbf A$ ascertain that the components of $\boldsymbol\nabla(A_{i,1}-A_{i,2})$ parallel to the surface are zero but not the component perpendicular to the surface? Is the component perpendicular to the surface different from those parallel to the surface? If so, how?

Then the author writes the normal component of $(10)$ is zero and the only non-zero component is the tangential component; I'm unable to comprehend why the tangential component is non-zero and not the normal component.

$\endgroup$
0
$\begingroup$

The field $\bf (A_1-A_2)$ has on the border due to the continuity of $\bf A$ a zero crossing on the border. Therefore the normal component of $\bf(A_1-A_2)$ can indeed have a slope while crossing. Therefore it is in the most general case non-zero. However, the component along the border (i.e. the tangential one) is always zero (you have the zero crossing there all along) , i.e. it does not change along the border. Therefore its derivative is zero. The reasoning of the textbook/pdf is correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy