How does one write the boundary condition for the electric field in this tidy package?

Question

The usual process for deriving the boundary condition for the tangential electric field across a surface uses Faraday's Law to prove that (when there are no dipole layers etc.) $$\hat{\textbf{t}} \cdot(\textbf{E}_1 - \textbf{E}_2) = 0 \tag{A}$$ where $\hat{\textbf{t}}$ is any vector in the tangent plane at the given point on the surface, and where 1 and 2 denote the two sides of the given surface. I have denoted this statement by $A$. If this is true, then

$$\hat{\textbf{n}} \times (\textbf{E}_1 - \textbf{E}_2) = 0 \tag{B}$$

follows immediately, for otherwise $\textbf{E}_1 - \textbf{E}_2$ is not completely along $\hat{\textbf{n}}$ (which is to say there is a component of $\textbf{E}_1 - \textbf{E}_2$ which is tangent to the plane, contradicting $A$).

Indeed, it is always written that $$A \iff B$$ and yet I cannot prove the converse, $B \implies A$. I'm hoping someone can supply that converse in full generality, without taking a particularly convenient surface etc.

Answer 1

$$\hat{\textbf{t}} \cdot(\textbf{E}_1 - \textbf{E}_2) = 0 \tag{A}$$ where $\hat{\textbf{t}}$ is any vector in the tangent plane at the given point on the surface, and where 1 and 2 denote the two sides of the given surface. I have denoted this statement by $A$. If this is true, then $$\hat{\textbf{n}} \times (\textbf{E}_1 - \textbf{E}_2) = 0 \tag{B}$$ follows immediately, for otherwise $\textbf{E}_1 - \textbf{E}_2$ is not completely along $\hat{\textbf{n}}$ (which is to say there is a component of $\textbf{E}_1 - \textbf{E}_2$ which is tangent to the plane, contradicting $A$).

Indeed, it is always written that $$A \iff B$$ and yet I cannot prove the converse, $B \implies A$.

Let $\vec v = \vec E_1 - \vec E_2$.

Your Eq (B) means that the angle between $\hat n$ and $\vec v$ is either zero degrees or 180 degrees. This means that $\vec v$ is either parallel to or anti-parallel to $\hat n$.

$$ \vec v = \pm |v|\hat n $$

Because, by definition, $\hat n$ is perpendicular to any $\hat t$, so too is $-\hat n$, and so too is $\vec v$.

$$ \vec v \cdot \hat t = \pm |v| \hat n \cdot \hat t = 0 $$