Page 268 - Electromagnetics

The transverse portion of the curl is merely

˜ ˜ ˜ ˜ ˜ ˜

˜
∂E z ∂E y ∂E x ∂E z ∂E z ∂E z
∇× E = ˆ x − + ˆ y − =−ˆ z × ˆ x + ˆ y
t
∂y ∂z ∂z ∂x ∂x ∂y
since the derivatives with respect to z vanish. The term in brackets is the transverse
˜
gradient of E z , where the transverse gradient operator is
∂
∇ t =∇ − ˆ z .
∂z
In circular cylindrical coordinates this operator becomes
∂ 1 ∂
ˆ
∇ t = ˆρ + φ . (4.211)
∂ρ ρ ∂φ
Thus we have

1
˜ ˜
H t (ρ,ω) = ˆ z ×∇ t E z (ρ,ω).
jω ˜µ
Similarly, the source-free Ampere’s law yields

1
˜ ˜
E t (ρ,ω) =− ˆ z ×∇ t H z (ρ,ω).
jω˜ c
These results suggest that we can solve a two-dimensional problem by superposition.
˜
˜
We first consider the case where E z = 0 and H z = 0, called electric polarization. This
case is also called TM or transverse magnetic polarization because the magnetic field is
transverse to the z-direction (TM z ). We have
1
˜
˜
2 ˜
2
(∇ + k )E z = 0, H t (ρ,ω) = ˆ z ×∇ t E z (ρ,ω). (4.212)
t
jω ˜µ
˜
Once we have solved the Helmholtz equation for E z , the remaining field components
˜
˜
follow by simple differentiation. We next consider the case where H z = 0 and E z = 0.
This is the case of magnetic polarization, also called TE or transverse electric polarization
(TE z ). In this case
1
˜
2 ˜
˜
2
(∇ + k )H z = 0, E t (ρ,ω) =− ˆ z ×∇ t H z (ρ,ω). (4.213)
t
jω˜ c
˜
˜
A problem involving both E z and H z is solved by adding the results for the individual
TE z and TM z cases.
Note that we can obtain the expression for the TE fields from the expression for the
TM fields, and vice versa, using duality. For instance, knowing that the TM fields obey
˜
˜
˜
˜
(4.212) we may replace H t with E t /η and E z with −ηH z to obtain
˜
E t (ρ,ω) 1
˜
= ˆ z ×∇ t [−ηH z (ρ,ω)],
η jω ˜µ
which reproduces (4.213).
© 2001 by CRC Press LLC