In electrostatics, one obtains the boundary conditions on the tangential components of $\vec E$ across two boundaries by using a small closed circuit so that \begin{align} \oint \vec E\cdot d\vec\ell =0 \qquad\Leftrightarrow \qquad \vec\nabla \times \vec E=0 \end{align} from which one deduces the continuity condition \begin{align} E_{1t}=E_{2t} \tag{1} \end{align}

The figure below is typical of the construct (here illustrated for the dielectric-dielectric boundary)

enter image description here

and taken from

Sadiku, Matthew N.O. Elements of Electromagnetics, Oxford University Press, 2014.

Variations of this can be found in most textbooks dealing with Maxwell's equations.

In the radiation regime, it is no longer true that $\vec E$ is irrotational since, because of induction \begin{align} \vec\nabla\times \vec E=-\frac{\partial \vec B}{\partial t} \tag{2} \end{align} yet the condition of Eq.(1) remain in use to obtain, for instance, the Fresnel equations.

Is there a derivation of (1) that works in the radiation regime where (2) holds and the induction cannot be neglected?

  • $\begingroup$ @hyportnex maybe you can expand and transform this in a detailed answer.. $\endgroup$ Feb 8, 2020 at 23:50

2 Answers 2


The integral form of Maxwell's equation, i.e., Faraday's law, where the flux and emf are calculated over a surface ${\mathcal S}$ and its boundary contour ${\mathcal{L}=\partial \mathcal S}$, resp., is $$\oint_{\mathcal{L}} \mathbf{E}\cdot d{\mathbf{l}} = -\frac{d}{dt}\iint_{\mathcal S} \mathbf{B}\cdot d\mathbf{s} \tag {1}$$ This $(1)$ is true for any surface ${\mathcal S}$ irrespective of whether the medium is continuous or not. Now let the contour $\mathcal{L}$ and its spanning surface $\mathcal{S}$ be the same as in your drawing $a-b-c-d-a$, and let $a-d$ and $b-c$ shrink to zero, then, normally, the flux will also be zero since you integrate the magnetic field over a vanishing area surface. Hence the line integral of $E$ over $d-c$ and $a-b$ are equal from which follows that $E_t$ is continuous.

Sometimes EEs introduce fictitious magnetic charges to make Maxwell's equations look the "same". This has nothing to do with magnetic monopoles, instead its only purpose is to appeal to analogies when calculating antennas that are basically the "negative" of metallic radiators, such as a horn or slot, a la Babinet. In the slot or horn a fictitious magnetic surface current is introduced that acts as the true source of radiation. The jump in the tangential $E_t$ is that fictitious surface current density. If you happen to know the radiation pattern of, say, a metal strip of the size of the slot in question then interchanging the $E$ and $H$ field you get the radiation pattern of the slot.

  • $\begingroup$ Ok fair enough but if you argue on the side of the contour you can also make $\Delta w$ arbitrarily small and thus obtain $(E_{1,t}-E_{2,t})\Delta w=0$ by effectively making $\Delta w=0$... or is your argument that the area of the small rectangle goes to $0$ as an infinitesimal of degree $2$ while $\Delta w\to 0$ as an infinitesimal of degree $1$? $\endgroup$ Feb 9, 2020 at 1:00
  • $\begingroup$ I would phrase it slightly differently, to me it is more the order of limits: fix a small $\Delta w$ so that the integral on, say, $a-b$ is approximately $E_{1t}\Delta w$, etc., and then let $\Delta h \to 0$. $\endgroup$
    – hyportnex
    Feb 9, 2020 at 1:08
  • $\begingroup$ I don’t see how the order of the limits enters. Maybe you are right... Can you provide a source? $\endgroup$ Feb 9, 2020 at 1:10
  • $\begingroup$ or maybe since area = $\Delta w\Delta h$ you can argue that the area goes to $0$ as $\Delta h\to 0$ even if $\Delta w$ is small... $\endgroup$ Feb 9, 2020 at 1:13
  • $\begingroup$ ... small but not zero yet, that is exactly how I see it $\endgroup$
    – hyportnex
    Feb 9, 2020 at 1:13

Consider a rectangular Amperian loop of sides $\Delta w$, $\Delta h$ (as per your right hand diagram), symmetrically placed with respect to the interface.

Faraday's law yields $$ (E_{1t} - E_{2t})\Delta w + \frac{1}{2}(E_{1n} + E_{2n})\Delta h - \frac{1}{2}(E_{1n} + E_{2n})\Delta h = -\frac{\partial B}{\partial t}\Delta w\Delta h,$$ where $E_{1n}$ and $E_{2n}$ are the normal components of the electric fields in medium 1 and medium 2.

$\Delta h$ can be made arbitrarily small, without changing the first term on the LHS, so let $\Delta h \rightarrow 0$. $$ (E_{1t} - E_{2t})\Delta w = 0$$

This is the treatment adopted in Electrodynamics (2017, 4th ed., Griffiths), p.343, though not written in explicit detail.


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