6

Electromagnetic Theorems

When an electromagnetic wave impinges on scattering objects, the incident

wave not only scatters off but also penetrates the scattering objects. If the
scattering objects are complex-shaped, the scattering analysis becomes com-
plicated; hence, electromagnetic theorems can be used to facilitate the problem
formulation. This chapter introduces some useful theorems and principles that
are often used in electromagnetic scattering analyses.

6.1 Uniqueness Theorem

When solving electromagnetic boundary-value problems, it is important to
ensure the uniqueness of the solutions. The uniqueness theorem provides the
boundary conditions that guarantee the uniqueness of solutions to boundary-
value problems. Consider the electromagnetic fields E and H within volume
V that contains the electric and magnetic sources Ji and M i . The problem
geometry is shown in Fig. 6.1. The uniqueness theorem can be stated as fol-
lows: The fields E and H can be uniquely determined if one of the following

Fig. 6.1. Bounded volume V containing sources Ji and Mi.

H. J. Eom, Electromagnetic Wave Theory for Boundary-Value Problems

© Springer-Verlag Berlin Heidelberg 2004
126 6 Electromagnetic Theorems

boundary conditions is specified on the surface S.

1. The tangential component fixE is given on S.

2. The tangential component fix His given on S.
3. The tangential component fi x E is given on a portion of S and the tan-
gential component fi x H is given on the rest of S.

Proof
Let us assume that two different fields (Ea, Ha) and (Eb, Hb) exist within
V. Each field must obey Maxwell's equations as

(6.1)

V X H a = -iweEa + Ji (6.2)

(6.3)

(6.4)

Subtracting (6.3) and (6.4) from (6.1) and (6.2), respectively, yields

V xEd =iwJ.tHd (6.5)

(6.6)

where Ed= Ea- Eb and Hd = Ha- Hb.

Let us consider

V · (Ed X H~) = ~ · (V X Ed) -Ed· (v x H~)

= ~ · (iwJ.tHd)- Ed· (iwe*~)

= iwJ.t 1Hdl 2 - iwe·j~j 2 (6.7)

where the symbol(·)* denotes the complex conjugate of(·).

Integrating (6.7) over V and utilizing the divergence theorem gives

[ V ·(Ed x H~) dv = t (Ed X H~) · ds

= [ iwJ.t IHdl 2 dv- [ iwe* 1Edl 2 dv . (6.8)
 6.2 Image Method 127

For analytic convenience, the medium is considered to be lossy with a complex

permeability J.l = J.lr + iJ.Li and complex permittivity € = fr + i~:i, where J.li and
fi account for medium loss. Substituting J.l and € into (6.8) finally produces

i (Ed X H~). as=- fv w (J.li 1Hdl 2 +~:i IEdn dv

+i [ w (J.Lr 1Hdl 2
- €r IEdn dv. (6.9)

Consider the vector identities

(6.10)

If one of the conditions 1, 2, or 3 is satisfied, then, either nx Ed = 0 or

n X H d = 0 on the surface s. Therefore
(Ed x H~) ·n = 0 (6.11)

indicating that the left-hand side of (6.9) becomes zero. The integrands of
the right-hand side of (6.9) are all real numbers. The only way to make the
right-hand side of (6.9) zero is to let

(6.12)

(6.13)

Therefore, as long as either condition 1, 2, or 3 is specified on the boundary

S, the field must be unique, such as (Ea = Eb and Ha = Hb)· This completes
the proof. A lossless medium is considered to be a limiting case of a lossy
medium with an infinitesimally small fi and J.li (~:i ---* 0 and J.li ---* 0), making
the uniqueness theorem also applicable to a lossless medium.

6.2 Image Method

The image method is a useful concept for solving problems associated with
infinitely extended PEC or PMC plane boundaries. This section introduces
the image method and its application to radiation from a small current above
a perfect electric conducting plane.

6.2.1 Image Method Using Infinite Planes

Let us consider the original problem in Fig. 6.2, illustrating the real sources
] and M above a PEC plane boundary of infinite extent. The real source
] represents a small electric current density, which is a small electric dipole
128 6 Electromagnetic Theorems

J J M M

real
sources
I e---0
~ ~ region (I)

original problem

M M

real
sources
I 0-------+0 ~ ~region (I)

I
image ~ ~gion (II)
sources

equivalent problem

Fig. 6.2. Image method with PEC (perfect electric conductor) plane. ffi: positive
electric charge, e: negative electric charge, -t: electric current density J, 83: positive
magnetic charge, B: negative magnetic charge, --~>: magnetic current density M.

consisting of two opposite electric point charges. Similarly, the real source
M represents a small magnetic current density, which is a small magnetic
dipole consisting of two opposite magnetic point charges. Since the outside
field produced by a small magnetic dipole is identical to the outside field
produced by a small electric current loop, a small magnetic dipole is considered
a small electric current loop. The total field in region (I) is the sum of the real
source contribution in the absence of a boundary and the reflection from the
boundary. The reflection is regarded as the response of an image source that
is placed in free space without a PEC boundary. The image sources must be
constructed so as to satisfy the tangential electric field continuity condition
 6.2 Image Method 129

requiring a zero tangential electric field at the boundary. Figure 6.2 illustrates
that the equivalent problem gives the same field in region (I) as the original
problem. The image method indicates that the total field in region (I) is the
response due to the real and imaginary sources.
As a dual problem, consider the real sources J and M that are placed
above the PMC boundary, as shown in Fig. 6.3. The image sources must
also be constructed so as to satisfy the tangential magnetic field continuity
requiring a zero tangential magnetic field at the boundary.

J J M M

real
sources 8----tB~a&a region (I)

original problem

J J M

real
sources
I 8----tB~a&a region (I)

I
image ,... ~-.... region (II)

8----tB'tl'~
sources

equivalent problem

Fig. 6.3. Image theorem with PMC (perfect magnetic conductor) plane. EB: positive
electric charge, e: negative electric charge, -+: electric current density J, EB: positive
magnetic charge, B: negative magnetic charge, ...... : magnetic current density M.
130 6 Electromagnetic Theorems

6.2.2 Current Above Perfect Electric Conducting Plane

z
E(r)

---- ------
I
h PEC
----
------d
X

Fig. 6.4. Small current above perfect electric conducting plane.

Consider an infinitesimally small magnetic current source M(r')

M(r') = zM c5(x')c5(y')&(z' -h) (6.14)
located above a perfect electric conductor, as shown in Fig. 6.4. Let us evaluate
the fields E(r) and H(r) radiating from M(r') when r »h. Based on the im-
age method, the electric vector potential F(r) is the sum of the contributions
from the real and image sources that are placed in free space. Substituting
M = zM&(x')c5(y')[&(z'- h)- c5(z' +h)] into a free-space solution (1.102)

_ f.l _
gives
eikir-r'i
F(r) = -4 M(r') I , dv'
1r V' r-r 1

(6.15)

where f. and k are the medium permittivity and wavenumber, respectively.

Note that rt and r2 are the distances from the real and image sources to the
observing point r, respectively. F(r) is expediently expressed using spherical
coordinates (r, (}, </J). Substituting the approximation in the radiation field at
r»h
 6.3 Equivalence Principle 131

r 1 ~r-hcosO {6.16)

r2 ~ r + hcosO {6.17)

into F(r) with z = f cos 0 - 0sin 0 yields

-
F(r) = - if.Meikr
21l"r
(
f cos 0 - 0 sin 0 sin( kh cos 0) .
A )

(6.18)

Substituting F(r) into

- 1 -
E= --v
f.
x F(r) (6.19)

and collecting the ~ terms gives

r
- A Mkeikr
E =¢ 21l"T sin 0 sin(kh cos 0) . {6.20)

Subsequently, substituting F(r) into

H = iwF(r) + _i_V [V. F(r)] (6.21)

WJ.Lf.

and collecting the ~ terms gives

r

H=
-

-Oy!I~E.p.
A

(6.22)

6.3 Equivalence Principle

The equivalence principle is a useful tool for solving boundary-value problems
in electromagnetic scattering and radiation. The equivalence principle allows
the actual sources to be replaced by equivalent, fictitious sources that produce
the same field within a region of interest. Some relevant discussion on the
equivalence principle is available in [7]. This section introduces Love's equiv-
alence principle and its application to the problem of transmission through a
circular aperture.

6.3.1 Love's Equivalence Principle

Let us consider problem {1) in Fig. 6.5, where the current sources] and M
radiate the fields (E, H) throughout regions (I) and (II). In problem {2), the
132 6 Electromagnetic Theorems

region (II)

problem (1)

region (II)
null field

problem (2)

Fig. 6.5. Equivalent problems.

sources ] and M are removed and the fields are assumed to be null in region
(I) and (E, H) in region (II), respectively. To retain a null field in region (I)
and ( E, H) in region (II), the surface currents ] s and M s must be placed on
the surface boundary
(6.23)

(6.24)

Then, the surface current densities ] 8 and M 8 radiate (E, H) in region (II),
while maintaining a null field in region (I). Problem (2) is thus equivalent to
problem (1) as far as region (II) is concerned.

6.3.2 Transmission Through Circular Aperture

Consider electromagnetic transmission through a circular aperture in a per-

fectly conducting plane, as shown in Fig. 6.6. The incident field impinges on
the aperture from below a perfectly conducting plane. Let us evaluate the
far-zone transmitted field for z > 0. For simplicity, the aperture electric field
 6.3 Equivalence Principle 133

F(r)

f
wave incidence

Fig. 6.6. Circular aperture in conducting plane.

is assumed to be Eap = xE0 • It is convenient to transform the original prob-

lem into an equivalent problem (b), as shown in Fig. 6.7, where the surface
magnetic current is M 8 (r1 ) = -n x Eap· The equivalent problem (b) in Fig.
6.7 can then be transformed into the equivalent problem (c) in Fig. 6.7, where
a PEC sheet is placed beneath the boundary S. The surface electric current
] 8 (r1 ) is short-circuited by the PEC sheet beneath the boundary S, and the

field (E, H) in region (II) is solely due to the surface magnetic current M 8 (r1 ).
According to the image method, the surface magnetic current M 8 (r1 ) on an
infinitely extended PEC can be considered as 2M 8 (r1 ) in free space, and the
electric vector potential for the radiated field for z > 0 is

F(r) = 4f { 2M 8 (r1 ) ~~~r~/

r- r
ds 1

1 _ 1_ ='I
7r } 8,

f ~ eik)r-r'l 1
= -4 ( -2n X Eap) ds

1 lr-
11" 8' r -r

EE eiklr-r'! d I
= - ~y -0 S. (6.25)
I
I
211" 8, r
Let us evaluate the electric vector potential F(r) when r »r 1• Note
lr-r 1~r-r1 cos'¢
1

= r - r 1 sinOcos(¢1 - ¢). (6.26)

134 6 Electromagnetic Theorems

f . 'dE
wave mCl ence

original problem (a)

z

2a
null field
equivalent problem (b)

PEC
equivalent problem (c)

--------~==*=~~---------+X
2a

equivalent problem (d)

Fig. 6. 7. Equivalent problems for circular aperture in conducting plane.

 6.3 Equivalence Principle 135

The electric vector potential F(r) is

-
F(r) ~
f.Eo
-y-
2 1l"
1s•
eikr
-
r
exp [-ikr' sinO cos(</>'-</>)] ds'

f.Eo eikr 1a 121T

= -y-
21l" - r o o
exp[-ikr'sin0cos(</> 1 -</>)]r1 d</>1 dr 1 .(6.27)

Substituting the identity

L
00

exp[-ikr'sinOcos(</>'-4>)] = inJn(-kr1 sinO)ein(t/l-.p) (6.28)

n=-oo

into F(r) results in

- eikr
F(r) = -yf.E0 -
1a J (-kr'sinO)r'dr
0 1 • (6.29)
r o
Since

y = rsinO sin 4> + 0cosO sin 4> +¢cos 4> (6.30)

J Jo(u)udu = uJ1(u) (6.31)

F(r) can be written as

F(r) = - (r sin 0 sin 4> + OcosO sin 4> +¢cos 4>)

·f.Eo eikr a2 J1 (ka sin 0)
r kasinO

(6.32)
- - - - 1
Let us evaluate E and H in the far zone. Substituting F(r) into E = --V' x
f.
F(r) and collecting the !r terms gives
- ik (
E = -€ -OF.p +</>Fe
A A )
(6.33)

Similarly, substituting F(r) into H = iwF(r) + _i_V' [Y' · F(r)] and col-
WJ.tf.
. t he -1 terms giVes
Iectmg .
r

(6.34)
136 6 Electromagnetic Theorems

6.4 Induction Theorem

The induction theorem is often used to solve electromagnetic scattering prob-
lems. This section first introduces the induction theorem, then considers its
application to scattering from a conducting rectangular plate.

6.4.1 Equivalence Based on Induction Theorem

• Consider an incident wave (Jt, It) that impinges on a dielectric scat-

terer, as shown in the original problem (1) in Fig. 6.8. The sources (J, M)
produce the incident field (Jt, F) in the absence of the scatterer. The
total field in region (II) consists of the incident and scattered components
(It, It) and ( E 8 , H 8 ) , respectively. The transmitted field in region (I)
is E ,H
(
-t) .
-t

problem (1)

problem (2)

Fig. 6.8. Equivalent problems.

 6.4 Induction Theorem 137

• Based on the induction theorem, the original problem (1) can be trans-
formed into the equivalent problem {2), where the field ( Et, Ht) is as-
sumed in region (I) and the field ( E 8 , H 8 ) is assumed in region (II).
At the boundary between regions (I) and {II), the surface currents ] 8 =
-t)
ft X ( -H8 - H = -ft X -=::iH and -M 8 = -ft X E - E = ft X -=i (-8 -t)
E must
be impressed in order to satisfy the boundary conditions.
• It is of interest to consider what happens when region (I) becomes a PEC
scatterer. Note that Et = Ht = 0 in region (I) and ft x + E8 ) = 0 (Jt
between regions (I) and (II). The impressed surface electric current ] 8 =
-ft x If is short-circuited on the PEC surface, and the field ( E 8 , H 8 ) in
region (II) is due solely to the impressed surface magnetic current M 8 =
nx -="E. .
A

6.4.2 Scattering from Conducting Rectangular Plate

Let us consider the problem of scattering from a perfectly conducting rectan-

gular plate based on the induction theorem. A uniform plane wave

Jt (x, y, z) = xEo exp [i(ky sin (}i - kz cos Oi)] (6.35)

is incident on an (a x b) perfectly conducting rectangular plate, as shown in
Fig. 6.9. Let us evaluate the far-zone scattered field for z > 0. If the size
of the plate is relatively large compared to the wavelength (a, b » A), it is
possible to assume that the plate is infinitely extended. According to the image
method, the surface magnetic current M 8 on an infinitely extended PEC can
be considered as 2M s in free space. The electric vector potential for z > 0 is
given by

F(r) = -4
€

~
1 ~
2ft x
.
E (x',y',o+) Ir-r 'I
eikir-r' 1
ds'

~ .!:..._ { 2ft
4~}~
X It (x', y 1, 0+) ~exp (ik(r- r 1 ·f)] ds 1
r

= t:eikr 1b/21a/2
4~r -b/2 -a/2

2ft X Jt (x', y 1, Q+)exp (-ik"f' ·f) dx' dy' . {6.36)

Let us evaluate the electric vector potential using spherical coordinates

(r, (}, ¢). Substituting
138 6 Electromagnetic Theorems

Ei z

~' I
I
"
"
n"

Fig. 6.9. Perfectly conducting rectangular plate of size (ax b).

n X 1lf (x', y', 0+) = yEo exp(iky' sin (}i) (6.37)

r' · f = (xx' + yy') · (x sin(} cos <P + y sin(} sin <P + z cos 0)

= x' sin (} cos <P + y' sin (} sin <P (6.38)

into F(r) and performing integration gives

F(r) = (r sin ll sin <P + 0cos(} sin <P +¢cos <P)

. Eeikr Eoab (sin a) (sin
/3
/3)
271'r a

(6.39)

where
ka . (} -~,
a= 2 sm cos'!' (6.40)

/3 = ~b (sin ll sin <P- sin Oi) (6.41)

 6.5 Duality Theorem 139

The far-zone scattered fields E and H are given by

- ik (
E= -€ -9Ft/J +¢Fe
A A )
(6.42)

H = iw (&Fe+ ¢Ft/J) . (6.43)

The far-zone fields (6.42) and (6.43) are good approximations of the exact
solution as long as (()i and 0) « ~ and (a and b) »A.

6.5 Duality Theorem

The duality theorem utilizes the symmetric property of Maxwell's equations.
Maxwell's equations for the electric source (J f:. 0 and M = 0) are given as

(6.44)

(6.45)

(6.46)

(6.47)

Similarly, Maxwell's equations for the magnetic source (M f:. 0 and J = 0)

are
(6.48)

(6.49)

(6.50)

(6.51)

A dual set for the electric and magnetic sources is shown in Fig. 6.10.
Maxwell's equations (6.48) through (6.51) can be constructed from (6.44)
through (6.47) by interchanging the quantities, as shown in Table 6.1 (Ee -t
Hm, He -t -Em, .. ·). Equations (6.44) through (6.47) and (6.48) through
(6.51) are termed dual to each other since they take a symmetric mathemat-
ical form. If the solution (Ee and He) to (6.44) through (6.47) is known,
140 6 Electromagnetic Theorems

/
.,...-- "
........

I M \
I \
I(- - ) I
\ \ E m1 Hm 1
J
I \ ' /
I (E H ) I '-.._ ___ .....
\ ., e I
''
----- /
/

Fig. 6.10. Dual set.

Table 6.1. Dual quantities

Electric source Magnetic source

Ee Hm

He -Em

J M

Pe Pm

p, f

f p,

A F

the solution to {6.48) through {6.51) can immediately be constructed by in-

terchanging the quantities, as shown in Table 6.1 (Ee -+ Hm, He -+ -Em,
... ) .
Consider radiation from a small electric current ] above a perfect magnetic
conductor plane, as shown in Fig. 6.11. The real current] above the PMC
surface is equivalent to the sum of the real] and its image source in a medium
of infinite extent. The problem considered in Fig. 6.11 is dual to the problem
considered in Fig. 6.4 in Section 6.2, and the radiation field in the far zone
 6.6 Reciprocity Theorem 141

.......... PMC
----..... ...............d

Fig. 6.11. Small electric current above perfect magnetic conductor plane.

is immediately given from the duality theorem. The solution to the problem
in Fig. 6.4 is given by {6.20) and {6.22). The solution to the dual problem in
Fig. 6.11 is thus
_ ~ keikr
H = -¢J-2- sin() sin(kh cos 0) {6.52)
-rrr
keikr
-o.,J-2--rrr
- A

E= sinOsin(khcosO) (6.53)

where.,= .J¥_.

6.6 Reciprocity Theorem

In circuit theory, the reciprocity theorem holds for any linear network. In
electromagnetic theory, the Lorentz reciprocity theorem holds for any linear
medium. The reciprocity theorem in circuit theory is a special form of the
Lorentz reciprocity theorem in electromagnetic theory. This section derives
the Lorentz reciprocity theorem, starting from Maxwell's equations.
142 6 Electromagnetic Theorems

Fig. 6.12. Two different sources and fields.

6.6.1 Lorentz Reciprocity Theorem

Consider two sets of Maxwell's equations in the same linear medium with
different sources. The problem geometry is shown in Fig. 6.12. Set (a) is given
by
{6.54)

{6.55)
Set (b) is
(6.56)

{6.57)

The vector operations \7 · (Ea x Hb) and \7 · (Eb x Ha) give

\7 · (Ea X Hb) = Hb · \7 X Ea- Ea · \7 X Hb

 6.6 Reciprocity Theorem 143

Subtracting {6.59) from {6.58) leads to

\7 · (Ea x Hb- Eb x Ha) =

The expression {6.60) is called the Lorentz reciprocity theorem in differential

form. Taking volume integration of {6.60) and utilizing the divergence theorem
on the left-hand side yields the Lorentz reciprocity theorem in integral form
as

i (Ea x Hb -Eb x Ha) ·as=

I[- (Ea · Jb- Ha · Mb) + (Eb · Ja- Hb · Ma)] dv. {6.61)

Two special cases of {6.61) are of practical interest. If region V is free of

sources (Ja = Ma = Jb = Mb = 0), {6.61) reduces to

{6.62)

If region V contains sources and V extends to infinity, the fields (Ea, Eb, Ha,
_ eikr
and Hb) take the asymptotic form- at r ~ oo. The fields have the(} and
T
¢ components that satisfy the relations, Es = rJHIP and EIP = -TJHs. Thus,
the left-hand side terms of {6.61) all vanish at r ~ oo, and {6.61) reduces to

6.6.2 Reciprocity for Antennas

Let us apply (6.63) to the problem of two small dipole antennas in free space,
as shown in Fig. 6.13. Assume that the electric current density Ja at antenna
1 produces Ea, which is the electric field at antenna 2. Similarly, the electric
current density Jb at antenna 2 produces Eb, which is the electric field at
antenna 1. Therefore, two antennas satisfy the Lorentz reciprocity theorem as

I Ea · Jb dv =I Eb · Ja dv. (6.64)

Let us reinterpret (6.64) in terms of the voltages and currents. The voltages
and currents are given as
144 6 Electromagnetic Theorems

antenna 1 antenna 2

Fig. 6.13. Two small dipole antennas.

J dl
v1 =- Eb. (6.65)

= JJa. as
11 (6.66)

%=-I dJ Ea· (6.67)

=I Jb ·as.
12 (6.68)

Therefore, the Lorentz reciprocity theorem (6.64) can be rewritten as

(6.69)

which amounts to the reciprocity theorem in circuit theory.

Problems for Chapter 6

1. Consider a small dipole electric current placed within a conducting
parallel-plate waveguide, as shown in Fig. 6.14. Construct the equivalent
problem based on the image method.
2. A uniform plane wave is normally incident on a thin PMC circular plate of
radius a. Assume a » ..\. Evaluate the backscattered far-zone field using
the induction theorem.
 Problems for Chapter 6 145

PEC

a t ~ B(x-x )8(y-y )O(z)

1 1

Fig. 6.14. Small dipole within conducting parallel-plate waveguide.

3. Consider radiation from a small electric current J above a perfect mag-

netic conductor plane, as shown in Fig. 6.11. Derive the solution (6.52)
and (6.53) without recourse to the duality theorem.