# Waveguides (Homework 1, Physics 323)

In class we found the “waveguide equation” $1/(λ_g^2 )+1/(λ_c^2 )=1/(λ_0^2 )$ , where $λ_g$ is the guided wavelength, $λ_c$ is the cutoff wavelength and $λ_0$ is the free-space wavelength for the simple waveguide consisting of two parallel mirrors.

a) Show that this relation also holds for the example of the rectangular waveguide

b) And thereby show that the group velocity for the rectangular waveguide is always less than or equal to the speed of light.

The waveguide equation is

$1/(λ_g^2 )+1/(λ_c^2 )=1/(λ_0^2 )$

a)

In a rectangular waveguide there can be TE (transversal electric) or TM (transversal magnetic) waves. For TE one has $E_z=0$, for TM one has $B_z=0$. (A TEM wave having simultaneously $E_z$ and $B_z=0$ is not possible). The discussion is given for TE waves. (For TM waves the discussion is similar).

Therefore if the waveguide is along the z direction and extends until x=a and y=b the longitudinal component of B is (from boundary conditions) (eq. 9.186):

$B_z (x,y)? 0$    and $B_z (x,y)=B_0*\cos⁡(k_x x)*\cos⁡(k_y y)$
with $k_x=mπ/a$      and $k_y=nπ/b$
where B_z needs to satisfy the equation (9.181 ii) (which is deduced directly from Maxwell equations written on components (x,y and z) of B and E):

$[∂^2/(∂x^2 )+∂^2/(∂y^2 )+(ω_0/c)^2-k_g^2 ] B_z (x,y)=0$

$ω_0$  is the frequency in free space

and $k_g$  and propagation wavevector in the guide

Or equivalent

$-k_x^2-k_y^2+ (ω_0/c)^2-k_g^2=0$

$(ω_0/c)^2=k_g^2+(k_x^2+k_y^2 )$    or  $ω_0^2=(ck_g )^2+c^2 π^2 [(m/a)^2+(n/b)^2]$
$ω_0^2=(ck_g )^2+ω_{mn}^2$     (1)
where $ω_{mn}$   is the cutoff frequency of the waveguide for transvesal mode (mn)

Since

$ω_0=ck_0$  ($k_0$ is the wavenumber in free space)

and $ω_{nm}=ck_{mn}$  ($k_{mn}=π\sqrt{(m/a)^2+(n/b)^2 )}$    one has
$(c k_0 )^2=(c k_g )^2+(c k_{mn} )^2$

And because the general definition (with or without indexes) $k=2π/λ$  one can write

$1/(λ_0^2 )=1/(λ_g^2 )+1/(λ_{nm}^2 )$

b)

From equation (1) one has

$k_g=1/c*\sqrt{ω_0^2-ω_{mn}^2}$

$k=1/c \sqrt{(ω^2-ω_{mn}^2 )}$
$v_g=(dω_0)/(dk_g )=dω/dk$     and from above  $dk=2ω*dω/c*1/(2\sqrt{(ω^2-ω_{mn}^2 )})$
$v_g=c/ω*\sqrt{(ω^2-ω_{mn}^2 )}=c\sqrt{(1-(ω_{mn}/ω)^2)}$
$ω_{mn}<ω$   and thus $v_g<c$