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On p.251 of Introduction to Electrodynamics by Griffiths, there is the formula for the discontinuity of the derivative of vector potential across a surface: $$ \frac{\partial \textbf{A}_{\text{above}}}{\partial n} - \frac{\partial \textbf{A}_{\text{below}}}{\partial n} = -\mu_0 \textbf{K}$$ I saw a way to prove this formula is to start from $$\textbf{B}_{\text{above}}-\textbf{B}_{\text{below}} = \mu_0(\textbf{K}\times \hat{\textbf{n}})$$ and take cross product with $\hat{\textbf{n}}$, \begin{align*} \hat{\textbf{n}} \times (\textbf{B}_{\text{above}}-\textbf{B}_{\text{below}}) &= \hat{\textbf{n}} \times (\nabla\times(\textbf{A}_{\text{above}} - \textbf{A}_{\text{below}}))\\ &= \nabla (\textbf{A}_{\text{above}}^\perp - \textbf{A}_{\text{below}}^\perp) - (\hat{\textbf{n}}\cdot \nabla)(\textbf{A}_{\text{above}} - \textbf{A}_{\text{below}}) \end{align*} whereas the RHS becomes $\mu_0 \textbf{K}$. We need to show $\nabla(\textbf{A}_{\text{above}}^\perp - \textbf{A}_{\text{below}}^\perp) = 0$, i.e. all three components are $0$. Suppose $z$ is the normal direction. One of the key step is to show that the $x,y$ derivative of any components of $\textbf{A}$ is continuous, but I don't see why this is true, since this cannot easily follow from the continuity of $\textbf{A}$. (I understand every step of the derivation after this). Does anyone have any idea on how to show the $x,y$ derivatives are continuous?

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You can assume $\nabla\cdot \vec{A} =0$.

Can you see that this is analogous to the no-monopole law of $\nabla \cdot \vec{B} =0$?

Therefore, just like the magnetic field, the component of the magnetic vector potential normal to the boundary is continuous and $\vec{A}^{\perp}_{\rm above} = \vec{A}^{\perp}_{\rm below}$.

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  • $\begingroup$ I understand that ${A}^{\perp}$ is continuous. But how are we gonna prove that its gradient is also continuous? $\endgroup$
    – Infinite
    May 8 at 7:27
  • $\begingroup$ @infinite what is the gradient of something that is always zero? $\endgroup$
    – ProfRob
    May 8 at 10:46
  • $\begingroup$ Sorry but I still didn't get your point. $\endgroup$
    – Infinite
    May 8 at 11:43
  • $\begingroup$ The $ \vec{A} $ itself is continuous at the boundary but still $\nabla\times( \vec{A}_{\text{above}}-\vec{A}_{\text{below}} )$ is non-zero. $\endgroup$
    – Infinite
    May 8 at 11:51

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