On a waveguide structure,such as coaxial cable,two wire lines,stripline,micro strip, the wave are propagating in Z direction.The E, H fields are of the form.
\begin{align*}
\pmb{E}&(x,y,z)=\pmb{E}(x,y)e^{-j\beta z} \\
\pmb{H}&(x,y,z)=\pmb{H}(x,y)e^{-j\beta z}
\end{align*}
For
source-free regions, maxwell equations become
\begin{align*}
\nabla & \times \pmb{E}= -j\omega \mu \pmb{H}\\
\nabla & \times \pmb{H}= j\omega \epsilon \pmb{E}
\end{align*}
Curl the above equations, using
\inline
\nabla \times \nabla \times \pmb{E}= \nabla ( \nabla \cdot \pmb{E})-\nabla^2 \pmb{E}
, we have
\begin{align*}
\nabla ^2& \pmb{E}+k^2 \pmb{E}=0 \\
\nabla ^2& \pmb{H}+k^2 \pmb{H}=0 \\
wh&ere ~ k=\omega \sqrt{\mu \epsilon}
\end{align*}
That's the helmholtz equations. Since both E and H has z-dependence in the form of
\inline -j\beta z ~ and ~ \partial_z^2= (-j\beta)^2=-\beta ^2 , the above equations reduce to
\begin{align*}
\nabla_T ^2& \pmb{E}(x,y)+k_c^2 \pmb{E}(x,y)=0 \\
\nabla_T ^2& \pmb{H}(x,y)+k_c^2 \pmb{H}(x,y)=0 \\
wh&ere ~ \nabla_T^2=\nabla_T \cdot \nabla_T=\partial_x^2+\partial_y^2 \\
an&d ~ k_c=k^2-\beta ^2
\end{align*}
The E/H and
\inline \nabla can be decomposed into longitudinal(z-direction) and transverse components(normal to z):
\begin{align*}
\pmb{E}&=\pmb{E}_T+\hat{z}E_z\\
\pmb{H}&=\pmb{H}_T+\hat{z}H_z\\
\nabla&= \hat{x}\partial_x+\hat{y}\partial_y+\hat{z}\partial_z=\nabla_T+\hat{z}\partial_z=\nabla_T-j\beta \hat{z}
\end{align*}
Separate Maxwell equation into transverse and longitudinal parts with the following identities
\begin{align*}
\hat{z} \times & \hat{z} =0 \\
\nabla_T & \times (\hat{z}E_z)=\nabla_T E_z \times \hat{z} \\
\nabla_T & \times (\hat{z}H_z)=\nabla_T H_z \times \hat{z} \\
\nabla_T & \times E_T ~ -> ~on ~ z-direction(\hat{x} \times \hat{y} =\hat{z}) \\
z \times & E_T ~ -> ~ on ~ T-plane (transverse plane) \\
\nabla_T & E_z \times \hat{z} ~ -> ~ on ~ T-plane \\
\end{align*}
Substitute
\inline \nabla = \nabla_T -j\beta\hat{z} and E=ET+zEz and H=HT+zHz into 1st maxwell equations,we have
\begin{align*}
\nabla_T & \times \pmb{E}_T = -j\omega \mu H_z \hat{z} ~~(1)\\
-j\beta & \hat{z} \times \pmb{E}_T +\nabla_T E_z \times \hat{z}= -j\omega \mu \pmb{H}_T ~~(2)
\end{align*}
Base on duality theorem, we have
\begin{align*}
\nabla_T & \times \pmb{H}_T = j\omega \epsilon E_z \hat{z} ~~(3) \\
-j\beta & \hat{z} \times \pmb{H}_T +\nabla_T H_z \times \hat{z}= j\omega \epsilon \pmb{E}_T ~~(4)
\end{align*}
Substitute z x(4) into (2), with identity
\hat{z} & \times \hat{z} \times A_T = -A_T \\
wh&ere ~ A_T \perp \hat{z}
We have
\begin{align*}
\pmb{H}_T&=-\frac{j\omega \epsilon}{k_c^2}\hat{z}\times \nabla_TE_z-\frac{j\beta}{k_c^2}\nabla_TH_z=-\frac{j\beta}{k_c^2}(\nabla_TH_z+\frac{1}{\eta_{TM}}\hat{z}\times \nabla_TE_z)...(5)\\
Ba&se ~ on ~ duality ~ theorem(E->H, H-> -E, \epsilon <-> \mu \eta <-> \frac{1}{\eta}) \\
\pmb{E}_T&=\frac{j\omega \mu}{k_c^2}\hat{z}\times \nabla_TH_z-\frac{j\beta}{k_c^2}\nabla_TE_z=-\frac{j\beta}{k_c^2}(\nabla_TE_z-\eta_{TE}\hat{z}\times \nabla_TH_z)...(6)\\
wh&ere ~~ k_c^2=k^2-\beta ^2=\omega ^2 \epsilon \mu -\beta ^2\\
\eta_{TM}&=\frac{\beta}{\omega \epsilon} \\
\eta_{TE}&=\frac{\omega \mu}{\beta}
\end{align*}
where k=w/c=w*sqrt(eu)=wavenumber in a propagation medium e,u. With Kc!=0, the transverse fields ET and HT can be solved with knowledge of Ez and Hz. To find a solution for Ez and Hz. usse reduced form of helmholtz equation(see above)
\begin{align*}
\nabla_T^2E_z+k_c^2E_z=0\\
\nabla_T^2H_z+k_c^2H_z=0
\end{align*}
TE mode (E field are purely transverse, no longitudinal component ) --
Ez=0
\begin{align*}
\pmb{E}_T&=\frac{j\beta}{k_c^2}\eta_{TE}\hat{z}\times \nabla_TH_z\\
\pmb{H}_T&=-\frac{j\beta}{k_c^2}\nabla_TH_z
\end{align*}
TM mode -- Hz=0
\begin{align*}
\pmb{E}_T&=-\frac{j\beta}{k_c^2}\nabla_TE_z \\
\pmb{H}_T&=-\frac{j\beta}{k_c^2}\frac{1}{\eta_{TM}}\hat{z}\times \nabla_TE_z
\end{align*}
TEM mode -- Ez=Hz=0.
For TEM mode, Kc=0. The problem is a electrostatic case, with
\begin{align*}
\nabla_T^2E_T=0\\
\nabla_T^2H_T=0
\end{align*}
Substitute Hz=0 into equ (1), we have
\begin{align*}
\nabla_T & \times \pmb{E}_T = 0 ~~ \Rightarrow ~~ E_T(x,y)= -\nabla_T \phi(x,y)\\
\nabla \cdot & \pmb{D}=0 ~~~ \Rightarrow ~~ \nabla_T \cdot E_T(x,y)= -\nabla_T ^2 \phi(x,y)=0 \\
\end{align*}
That's electrostatic field problem.
Impedance relationship
\begin{align*}
\eta &=\sqrt{\mu /\epsilon}\\
\beta &= \sqrt{k^2-k_c^2}=k\sqrt{1-k_c^2/k^2}=\frac{\omega}{c}\sqrt{1-\omega_c^2/\omega^2}\\
\eta &_{TE}=\frac{\omega \mu}{\beta}=\frac{\eta}{\sqrt{1-\omega_c^2/\omega^2}}\\
\eta &_{TM}=\frac{\beta}{\omega \epsilon }=\eta \sqrt{1-\omega_c^2/\omega^2}
\end{align*}
In cylindrical coordinate,
\pmb{E}_T=\hat{\rho}E_{\rho}+\hat{\phi}E_{phi},
\begin{align*}
\nabla_T=\hat{\rho}\frac{\partial}{\partial \rho}+\hat{\phi}\frac{\partial}{\rho \partial \phi}
\end{align*}
