# Metallic Waveguide Boundary

I've been studying TE and TM modes on waveguides for microwave frequencies.

It basically consists on applying Maxwell equations on a rectangular metallic cavity, with infinite length on $z$ direction, and finite on $x$ and $y$ directions ( say, length $a$ on $x$ and $b$ on $y$).

On TM modes, we admit $H_z$ =0, and apply boundary conditions $E_z$= 0 on $x= 0,x=a,y=0,y=b$ ( since its a conductor )

On TE modes,we admit $E_z$=0, and apply boundary conditions $\displaystyle\frac{\partial H_z}{\partial y} = 0$ on $y=0,y=b$ and $\displaystyle\frac{\partial H_z}{\partial x} = 0$ on $x=0,x=a$.

My question is: why TE boundary conditions involve derivatives of $H_z$ equal to zero, and not $H_z$ itself null?

All of these calculations can be found on Sadiku's Electromagnetism ,page 496 if 3rd edition.

Maxwell's Equation: $$\\mathbf{\nabla} \times \mathbf{H} = \epsilon \frac{\partial \mathbf{E}}{\partial t}\$$

Expanding and using $$\E_z=0\$$ for TE, we get: $$\\frac{\partial H_z}{\partial y}\$$ gives $$\E_x\$$, and $$\\frac{\partial H_z}{\partial x}\$$ gives $$\E_y\$$.

Boundary conditions require: electric field cannot be parallel to metal surface.

$$\\mathbf{E_x}\$$ is parallel to top & bottom walls, i.e. y=0 and y=b. $$\\mathbf{E_y}\$$ is parallel to left & right walls, i.e. x=0 and x=a.

So $$\E_x\$$, hence $$\\frac{\partial H_z}{\partial y}\$$, should be 0 at y=0 and y=b. And, $$\E_y\$$, hence $$\\frac{\partial H_z}{\partial x}\$$, should be 0 at x=0 and x=a.

"why... not Hz itself null?"

TE, by definition, demands that $$\E_x\$$ and $$\E_y\$$ may be nonzero, $$\E_z\$$ is zero, and $$\H_z\$$ is nonzero.

• I appreciate your time and knowledge, thank you so much! – Mateus Mar 7 at 22:30