Why is the curl of electric field vector inside a hollow waveguide zero for TEM waves?

Question

I have been studying David Griffiths 'Introduction to Electrodynamics', Electromagnetic Waves chapter. For hollow wave guide, the Maxwell's equations

$$\begin{align} \nabla\cdot\mathbf E &= 0 \\ \nabla\times\mathbf E &= - \frac{\partial \mathbf B}{\partial t} \end{align}$$

From these we can obtain,

$$\begin{align} \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= i\omega B_z \\ \frac{\partial E_z}{\partial y} - ikE_y &= i\omega B_x \\ ikE_x - \frac{\partial E_z}{\partial x} &= i\omega B_y \end{align}$$

The left hand side of these equations are the components of the curl of the electric field vector. To show that TEM waves (where the z component of both the electric and magnetic fields are zero) cannot exist in a hollow wave-guide, it's stated that, the divergence and curl of the electric field vector are both zero. My question is, how can we show that the curl is zero for this electric field?

Answer 1

I assume that Griffiths does not mean the curl of the total electric field, just the transverse part of the electric field.

If we assume a wave of the following kind $$\mathbf{E} = \mathbf{E}_T(x,y)e^{j(\omega t - kz)},$$

and we define a tranverse del operator as $\nabla_T = \hat{x}(\frac{\partial}{\partial x}) + \hat{y}(\frac{\partial}{\partial y}),$ then we can state that $\nabla_T \cdot \mathbf{E}_T = 0 $ and $\nabla_T \times \mathbf{E}_T = 0$. The tranverse field can thus be written in terms of a potential function, similar to the electrostatics case. It is solved by formulating Laplace's equation with the appropriate boundary equations.