My textbook says that

The laws of Electromagnetic Theory (Section 3.1) lead to certain requirements that must be met by the fields, and they are referred to as the boundary conditions. Specifically, one of these is that the component of the electric field E that is tangent to the interface must be continuous across it (the same is true for H). In other words, the total tangential component of E on one side of the surface must equal that on the other (Problem 4.37). Thus, since un is the unit vector normal to the interface, regardless of the direction of the electric field within the wavefront, the cross-product of it with Un will be perpendicular to un and therefore tangent to the interface.

So the incident, reflected, and transmitted waves must have their components that are in the plane of the boundary add up in this way:

$$\mathbf{u_{n}}\times\mathbf{E_{i}+u_{n}\times E_{r}=u_{n}\times E_{t}}$$

This is supposed to keep things continuous. I understand that in order for a function to be differentiable it can't have any discontinuities, but it's not clear how that comes into play here. Why the boundary and not at all times? I would super appreciate if someone could just throw a definition or an example that would help to explain how this works.

  • 2
    $\begingroup$ "Why the boundary and not at all times?" Where else do you have reflection? (The answer of course is there can be reflection when there is a gradient in the appropriate material properties, but you handle that as a special case of boundaries, so treat the question in terms of discrete changes...) $\endgroup$ – dmckee Jan 30 '16 at 0:56
  • $\begingroup$ Any homogeneous medium can be trivially viewed of a boundary of the medium with itself. The boundary conditions are satisfied by mere leaving out any reflected wave in such a case -- formally, the boundary conditions thus hold in any point of the whole space, just as Maxwell equations do. $\endgroup$ – dominecf Feb 3 '16 at 9:06

Your Answer

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

Browse other questions tagged or ask your own question.