TEM waves in waveguides and free space

Question

I have seen the mathematical explanation as to why TEM waves cannot exist in hollow waveguides (eg: https://www.youtube.com/watch?v=G8u2WEBF7MY), But that derivation holds good for plane waves in free space as well. So, does that mean plane waves doesn't exist as well? But all the basic books in optics start with a planar wave for wave optics. What am I missing here?

Answer 1

In its simplest form (homogeneous propagation medium), both cases are described dynamically by the Helmholtz wave equation (https://en.wikipedia.org/wiki/Helmholtz_equation): $$\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}+\frac{\partial ^2 u}{\partial z^2}-\frac{1}{c^2}\frac{\partial ^2 u}{\partial t^2}=0 \tag{1}\label{1}$$ In $\eqref{1}$ the function $u=u(x,y,z,t)$ may represent any component of the $H$ or $E$ or that of their potentials.

To solve such equation one must add certain boundary and initial conditions, otherwise the solution of the partial differential equation is not defined. Usually, in EM, we avoid specifying the initial conditions by assuming that the solution can be represented by its Fourier transform, and we care about the stationary harmonic solutions that satisfy homogeneous boundary conditions that we will impose.

For propagation in free space the functions $e^{\mathfrak j (\omega t - \mathbf{k \cdot r)}} $ do solve the Helmholtz equation $\eqref{1}$ and if we do not impose anything else on the solution's behavior at infinity then $\hat u(\mathbf{k}) = A(\mathbf{k}) e^{\mathfrak j (\omega t- \mathbf{k \cdot r})} $ is a general solution for arbitrary $\mathbf{k}$ vector and $\omega$ as long as the dispersion condition holds $$c^2 = \frac{\omega^2}{\mathbf{k}^2}\tag{2}\label{2}.$$ Consequently, all free space solutions can be represented as the Fourier Integral $$u(\mathbf r, t) = e^{\mathfrak j \omega t}\int\int\int_{-\infty}^{\infty} \hat u(\mathbf{k}) e^{- \mathfrak j\mathbf{k \cdot r}} d^3\mathbf{k}$$

If you constrain your wave into a waveguide, you impose additional boundary conditions on the wave that does not exist om free space. For example, for a circular metal pipe of radius $a$ lying along the $z$ axis the E field must be such that when written in cylindrical coordinates $[r, \theta, \varphi]$: $E_{\varphi}(r=a, \theta,\varphi) = 0$ and $H_r(r=a, \theta,\varphi)=0.$ Now an infinite long pipe is not a bounded system but if consider only homogeneous waves along the $z$ axis then we can write the solutions as a product of a harmonic term in $z$ and another one that satisfies the 2D Helmholtz equation: $u=e^{-\mathfrak j \beta z } v(x,y)$ and $\partial_x ^2 v + \partial_y ^2 v +(\mathbf k ^2 - \beta^2)v=0$. Now all boundary constraints are imposed on the function $v$: $v(r=a)=0$ or $\partial_r v|_{r=a}=0$. The 2D Helmholtz equation has the well-known TE or TM solutions but as $\beta \to 0$ both sets of solutions tend to zero in conformance with Dirichlet's theorem that a 2D harmonic function in a singly connected domain is zero if it is zero at the boundary.

On the other hand, if the domain is multiply connected, e.g., an annulus (coaxial pipe), then the function does not have to be zero. In fact, if we start with prescribing the so-called scalar potential at the boundaries to be two unequal constants we get the TEM mode, and similarly we can have multiple conductors coupled that way and get a variety of TEM-like propagation by prescribing the scalar potential to be constant on each metal. It is conventional to set one metal at "zero" potential, but it is not necessary. Notice that the spatial distribution of the modes are not uniform unlike that of the free space harmonics, and, of course the difference is caused by the very different boundary conditions, none specific for free-space and very restrictive, that is constant, for the pipes.