In Griffiths's Introduction to Electrodynamics, monochromatic guided waves are proposed to have the form

$$\mathbf{\tilde{E}}(x,y,z,t)=\mathbf{\tilde{E}}_0(x,y)e^{i(kz-\omega t)}$$ $$\mathbf{\tilde{B}}(x,y,z,t)=\mathbf{\tilde{B}}_0(x,y)e^{i(kz-\omega t)}$$


$$\mathbf{\tilde{E}}_0=E_x\mathbf{\hat{x}}+E_y\mathbf{\hat{y}}+E_z\mathbf{\hat{z}}$$ $$\mathbf{\tilde{B}}_0=B_x\mathbf{\hat{x}}+B_y\mathbf{\hat{y}}+B_z\mathbf{\hat{z}}$$

Then the following is stated:

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In every denominator the expression $(\omega /c)^2-k^2$ appears. But, as far as I know, $\omega /c=k$, so

$$(\omega /c)^2-k^2=k^2-k^2=0$$

What am I missing here?

  • $\begingroup$ Copy-paste from the 3rd Edition : Problem 9.27 Show that the mode $\:\mathrm{TE}_{00}\:$ cannot occur in a rectangular wave guide. [ Hint : In this case $\:\omega/c=k\:$, so Eqs. 9.180 are indeterminate, and you must go back to 9.179. Show that $\:B_{z}\:$ is a constant, and hence-applying Faraday's law in integral form to a cross section-that $\:B_{z}=0\:$, so this would be a $\:\mathrm{TEM}\:$ mode.] $\endgroup$
    – Frobenius
    Jan 17, 2017 at 6:06
  • $\begingroup$ I don't if it is the case, but in some books they refer to $k$ as the vector of the wave in a certain medium, therefore being equal to $k=\frac{ωn}{c} $ with $n$ refractive index of that medium. Does it make sense? $\endgroup$
    – JackI
    Jan 17, 2017 at 6:43

2 Answers 2


$\omega=kc$ for plane waves which have spatial variation only along $k$. In a waveguide, the field varies in other directions as well and the relation is incorrect.

You can think about a waveguide mode as being composed of a standing waves pattern in the plane of the waveguide combined with the propagation along the normal direction. If you found the total wavenumber $k_0$ including these standing wave components as well, you would get the $\omega=k_0c$ relation. Since $k$ here only corresponds to the propagating part of the wave vector and not the total wavenumber the relation $\omega=kc$ does not hold.


Usually, people don't use the symbol $k$ in this context so as to avoid exactly the kind of confusion you are having.

In this context, the symbol written as $k$ in Griffiths's equations is often written $\beta$ or $k_z$; it is then called the propagation constant and it depends on the geometry of the wave in the waveguide. It is found through an eigenvalue equation that arises from the waveguide's boundary conditions. The discrete spectrum of the eigenvalue equation defines the bound modes of the waveguide.

$\beta$ is always less than the wavenumber for the medium within the waveguide, so that you don't get singularities in Griffiths's equations.

For example, in a two dimensional waveguide comprising an air channel (or "slab") between two perfectly conducting walls, the modes are actually the superpositions of two plane waves propagating in pairs at angles $\pm\theta_j$ (one is the other reflected off the walls in accordance with the law of reflexion). So $\beta_j = k\,\cos\theta_j$ and the angles $\theta_j$ are defined by the boundary condition that the longitudinal component of the electric field must vanish at the walls in a perfectly conducting waveguide.


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