References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.27.

The wave equations for an electromagnetic wave in a wave guide are

If we have a *transverse electric* or TE wave, so we need to solve only the equation. For a rectangular wave guide with dimensions of in the direction and in the direction, we can use separation of variables to get a solution (the technique is mathematically the same as that used in solving the infinite square well in quantum mechanics). That is, we assume that

and substitute this into 2, then divide through by :

Because this equation must hold for all values of and , the two derivative terms must separately be constant, so we have

for some constants and , which satisfy

The general solution to these ODEs can be written as either trigonometric functions or complex exponentials. If we choose trig functions, we have

The boundary conditions require that at and

The component of perpendicular to the walls of the guide in the direction is , and since inside the conducting wall, this condition requires . In our derivation of the wave equation for a wave guide, we found that

and since , we must also have , which in turn requires that

From 8, this means that and

so

for equal to a non-negative integer.

Exactly the same analysis on gives us

so the separation of variables solution gives us

This is known as the mode. From 7 we get the wave number

In fact, at least one of or must be non-zero, as we can see by the following argument. Suppose . Then . We need the results we got in the previous post:

With and we get from 21 and 19

Adding these equations gives us

Similarly, from 22 and 18 we get

Therefore, is constant in both the and directions, so it is a constant overall. To find what constant this is, we can use Faraday’s law in integral form:

The area of integration on the RHS that we’ll choose is a cross-section of the wave guide, so the path of integration on the LHS is around the rectangular boundary of the guide. We know that

so

In this case points in the direction, so we get

since is constant over the area of integration.

Since we’ve chosen the path of integration to be the boundary of the wave guide, we can use the boundary condition

The field inside the conducting wall of the wave guide is zero, so the parallel field at the boundary must also be zero and, since , the line integral must also be zero, so we must have

That is, if , both the electric and magnetic fields must be transverse (known as TEM mode). Griffiths shows in section 9.5 of his book that in this case, for a hollow wave guide of any cross section, the electric field must actually be zero everywhere, which means no wave can propagate down the guide. Thus the mode cannot exist in a hollow wave guide.