Page 391 - Electromagnetics

Nowwe are ready to give a longitudinal–transverse decomposition of the fields in a
lossy, homogeneous, isotropic region in terms of the direction ˆ u. We write Maxwell’s
equations as
˜
˜
˜
˜ i
∇× E =− jω ˜µH t − jω ˜µˆ uH u − J ˜ i − ˆ uJ , (5.103)
mt mu
˜ i
c ˜
˜
˜ i
c ˜
∇× H = jω˜ E t + jω˜ ˆ uE u + J + ˆ uJ , (5.104)
t u
where we have split the right-hand sides into longitudinal and transverse parts. Then,
using (5.99) and (5.100), we can equate the transverse and longitudinal parts of each
equation to obtain
˜
˜
˜ i
∇ t × E t =− jω ˜µˆ uH u − ˆ uJ , (5.105)
mu
˜
˜ ∂E t ˜ ˜ i
− ˆ u ×∇ t E u + ˆ u × =− jω ˜µH t − J , (5.106)
mt
∂u
c ˜
˜
˜ i
∇ t × H t = jω˜ ˆ uE u + ˆ uJ , (5.107)
u
˜
˜ ∂H t c ˜ ˜ i
− ˆ u ×∇ t H u + ˆ u × = jω˜ E t + J . (5.108)
t
∂u
We shall isolate the transverse fields in terms of the longitudinal fields. Forming the
cross product of ˆ u and the partial derivative of (5.108) with respect to u, we have
˜ 2 ˜ ˜ ˜ i
∂ H u ∂ H t c ∂E t ∂J t
−ˆ u × ˆ u ×∇ t + ˆ u × ˆ u × 2 = jω˜ ˆ u × + ˆ u × .
∂u ∂u ∂u ∂u
Using (B.7) and (B.80) we find that
˜ 2 ˜ ˜ i
∂ H u ∂ H t c ∂E t ∂J t
∇ t − 2 = jω˜ ˆ u × + ˆ u × . (5.109)
∂u ∂u ∂u ∂u
c
Multiplying (5.106) by jω˜ we have
˜
c c ∂E t 2 c ˜ c ˜ i
˜
− jω˜ ˆ u ×∇ t E u + jω˜ ˆ u × = ω ˜µ˜ H t − jω˜ J . (5.110)
mt
∂u
˜
We nowadd (5.109) to (5.110) and eliminate E t to get

2 ˜ ˜ i
∂ 2 ∂ H u c c ˜ i ∂J t
˜
˜
+ k H t =∇ t − jω˜ ˆ u ×∇ t E u + jω˜ J mt − ˆ u × . (5.111)
∂u 2 ∂u ∂u
This one-dimensional Helmholtz equation can be solved to find the transverse magnetic
˜
˜
field from the longitudinal components of E and H. Similar steps lead to a formula for
˜
the transverse component of E:

2 ˜ ˜ i
∂ 2 ∂E u ∂J mt
˜
˜ i
˜
+ k E t =∇ t + jω ˜µˆ u ×∇ t H u + ˆ u × + jω ˜µJ . (5.112)
t
∂u 2 ∂u ∂u
˜
˜
We find the longitudinal components from the wave equation for E and H. Recall that
the fields satisfy
1
˜ i
˜ i
2 2 ˜ i
(∇ + k )E = ∇ ˜ρ + jω ˜µJ +∇ × J ,
m
˜ c
1
2 2 ˜ i c ˜ i
˜ i
(∇ + k )H = ∇ ˜ρ + jω˜ J −∇ × J .
m
m
˜ µ
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