Zdunkowski W., Trautmann T., Bott A. Radiation in the atmosphere: A course in Theoretical Meteorology
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9.2 Maxwell’s equations 335
where we have introduced the square of the wave number
k
2
= ω
2
εµ +
iµσ
ω
(9.6)
The following quantities will also be needed in our derivation:
(9.7)
speed of light in vacuum: c =
1
√
ε
0
µ
0
speed of light in a nonconducting medium: v =
1
√
εµ
wave number in vacuum: k
0
=
ω
c
wave number in a nonconducting medium: k =
ω
v
complex index of refraction: N = c
&
εµ +
iµσ
ω
The latter equation follows from (8.253). The imaginary part of N indicates that
absorption is taking place. From (9.6) and (9.7) one may easily see that
k
2
= k
2
0
N
2
(9.8)
From (9.7) it is easily seen that in a dielectric medium with σ =0 the wave number
reduces to k =ω/v. In a nonabsorbing medium the imaginary part of N vanishes,
that is κ =0 in (8.253). The last equation of (9.7) leads to the conclusion that
every conductor absorbs electromagnetic waves while every dielectric (insulator)
is transparent. Practical examples can be given to contradict this statement. The
reason is that not only the refractive index depends on frequency but also the
conductivity.
Utilizing (9.8) we may also write (9.5) as
∇
2
E + k
2
0
N
2
E = 0, ∇
2
H + k
2
0
N
2
H = 0
(9.9)
For these two differential equations we assume the trial solutions
E = E
0
exp
[
i(k · r − ωt)
]
, H = H
0
exp
[
i(k · r − ωt)
]
(9.10)
where k and r are the wave number and position vectors. If e
k
is a unit vector in
the direction of wave propagation then the wave number vector is given by
k = ke
k
= k
0
N e
k
(9.11)
336 Light scattering theory for spheres
n
2
n
1
n
∆ A
r
∆ h
S
ε
2
,µ
2
, σ
2
ε
1
,µ
1
, σ
1
Fig. 9.1 Illustration for the derivation of the normal boundary conditions for B
and D.
9.3 Boundary conditions
The field equations are valid for ordinary points in space in whose neighborhood
the field vectors change continuously. Across any surface separating two media
sharp changes may occur in the parameters characterizing the media so that the
field vectors are expected to exhibit corresponding changes. Let these parameters
be denoted by ε
1
,µ
1
,σ
1
and ε
2
,µ
2
,σ
2
.
For the solution of the wave equations suitable boundary conditions must be
supplied. The boundary surface separating the two media is denoted by the symbol
S and has the normal unit vector n, see Figure 9.1. We imagine a very thin transition
layer in which the parameters change rapidly but continuously from their values
near S in (1) to their values near S in (2). Within this transition layer as well
as in (1) and (2) the field vectors and their first derivatives are assumed to be
continuous bounded functions of position and time. Through the transition layer
we now construct a small right cylinder of height h and cross-section area A.
The surface normal unit vectors n
1
, n
2
at the bottom and the top of the cylinder
have the directions shown in the figure.
Integrating the divergence equations of (9.2) over the volume V = hA of
the cylinder and utilizing the divergence theorem of Gauss yields
V
∇·BdV =
0
B · dA = 0
V
∇·DdV =
0
D · dA = q =
V
ρdV
(9.12)
where the integrals over dA are taken over the closed cylinder surface. For
the evaluation of the surface integrals we assume that with vanishing height
9.3 Boundary conditions 337
t
2
t
1
n
n
A
t
S
∆ h
A
∆s
∆s
ε
2
,µ
2
, σ
2
ε
1
,µ
1
, σ
1
Fig. 9.2 Illustration for the derivation of the tangential boundary conditions for B
and D.
h → 0 of the cylinder the contribution of the walls may be ignored. Thus we
may write
0
B · dA =
0
A
B
1
· n
1
dA+
0
A
B
2
· n
2
dA = 0
0
D · dA =
0
A
D
1
· n
1
dA+
0
A
D
2
· n
2
dA =
¯
qA
(9.13)
Here we have introduced the surface charge density
¯
q = q/A = ρh which is
the charge of the cylinder per unit area. We observe that the total charge q remains
constant since it cannot be destroyed. As h → 0 the volume charge becomes
infinite.
Since n
2
=−n
1
= n, see Figure 9.1, we have
0
A
(
B
2
− B
1
)
· ndA = 0,
0
A
(
D
2
− D
1
)
· ndA =
¯
qA (9.14)
Assuming that the field vectors are constant within a small area element A we
finally obtain the boundary conditions for the normal components of B and D
(
B
2
− B
1
)
· n = 0,
(
D
2
− D
1
)
· n =
¯
q
(9.15)
These equations state that the normal component of B across any discontinuity
surface is continuous. In contrast, the normal component of D experiences an abrupt
change whose magnitude is
¯
q.
In order to express the behavior of the tangential components of B and D at the
discontinuity surface S we consider a closed surface A with normal unit vector n
A
,
see Figure 9.2. The tangential unit vectors along the lower and upper boundary of
A are denoted by t
1
and t
2
.
338 Light scattering theory for spheres
Integrating (9.1) over this surface yields together with (9.2)
A
∇×E · n
A
dA =
0
E · ds =−
A
∂B
∂t
· n
A
dA
A
∇×H · n
A
dA =
0
H · ds =
A
∂D
∂t
+ J
· n
A
dA
(9.16)
where by means of the Stokes integral theorem in each equation the first surface
integrals have been rewritten as closed line integrals.
We split the line integrals into four parts and assume that in the limit h → 0
the contributions from the corresponding sides vanish, see Figure 9.2. Hence we
obtain
s
E
1
· t
1
ds +
s
E
2
· t
2
ds =−
A
∂B
∂t
· n
A
dA
s
H
1
· t
1
ds +
s
H
2
· t
2
ds =
A
∂D
∂t
+ J
· n
A
dA
(9.17)
For a small surface A we have t
2
=−t
1
= t = n
A
× n so that
s
(
E
2
− E
1
)
· n
A
× nds =−
A
∂B
∂t
· n
A
dA
s
(
H
2
− H
1
)
· n
A
× nds =
A
∂D
∂t
+ J
· n
A
dA
(9.18)
By assuming that for small values of A all integrands are constant, we obtain from
(9.18)
(
E
2
− E
1
)
· n
A
× n =−
∂B
∂t
· n
A
h
(
H
2
− H
1
)
· n
A
× n =
∂D
∂t
+ J
· n
A
h
(9.19)
For finite values of the current J the right-hand sides of these equations vanish
in the limit h → 0 since the field vectors B and D and their derivatives with
respect to time remain bounded. Thus, we finally obtain the result that the tangential
components of the field vectors are continuous across a discontinuity surface S, i.e.
n ×
(
E
2
− E
1
)
= 0, n ×
(
H
2
− H
1
)
= 0
(9.20)
Equations (9.15) and (9.20) describe the boundary conditions which are neces-
sary to solve the differential equations (9.9). We will apply these equations to the
scattering of radiation by a sphere (medium 2) which is embedded in the atmosphere
(medium 1). The boundary between the two media is the spherical surface of the
scatterer. Thus, it will be expedient to state the components of the field vectors in
spherical coordinates.
9.4 The solution of the wave equation 339
9.4 The solution of the wave equation
9.4.1 Solution of the scalar wave equation in spherical coordinates
The solution of the scalar form of the wave equation is instrumental in the con-
struction of the solution of the vector wave equation. Given is the scalar form of
the wave equation for the scalar quantity u(x, y, z, t)
∇
2
u =
1
c
2
∂
2
u
∂t
2
(9.21)
For this equation we wish to express the solution in spherical coordinates. Assuming
that u can be written in the form
u(x, y, z, t) = u
∗
(x, y, z) exp(−iωt ) (9.22)
we find
∇
2
u
∗
+ k
2
u
∗
= 0 with k =
ω
c
(9.23)
and in spherical coordinates
∂
2
u
∗
∂r
2
+
2
r
∂u
∗
∂r
+
1
r
2
sin ϑ
∂
∂ϑ
sin ϑ
∂u
∗
∂ϑ
+
1
r
2
sin
2
ϑ
∂
2
u
∗
∂ϕ
2
+ k
2
u
∗
= 0 (9.24)
Defining the angular part of the Laplacian by
∇
2
1
u
∗
=
1
sin ϑ
∂
∂ϑ
sin ϑ
∂u
∗
∂ϑ
+
1
sin
2
ϑ
∂
2
u
∗
∂ϕ
2
(9.25)
the wave equation can be written as
∂
2
u
∗
∂r
2
+
2
r
∂u
∗
∂r
+
1
r
2
∇
2
1
u
∗
+ k
2
u
∗
= 0 (9.26)
We assume that the solution to (9.26) can be written in the form
u
∗
(r,ϑ,ϕ) = f (r)F(ϑ, ϕ) (9.27)
Substitution of (9.27) into (9.26) yields
r
2
f
d
2
f
dr
2
+
2r
f
df
dr
+ k
2
r
2
=−
1
F
∇
2
1
F (9.28)
By the method of separation of the variables each side must be equal to a con-
stant C. In order to guarantee unique solutions of these equations we set C =
n(n + 1) with n = 0, 1,...For details see, for example, Smirnow (1959). Thus we
340 Light scattering theory for spheres
obtain
(a) ∇
2
1
F + CF = 0
C = n(n + 1), n = 0, 1,...
(b)
r
2
f
d
2
f
dr
2
+
2r
f
df
dr
+ k
2
r
2
− C = 0
(9.29)
The solution to (9.29a) is well-known and is given by
F
n
(ϑ, ϕ) = a
n
P
n
(cos ϑ) +
n
m=1
[
a
mn
cos(mϕ) + b
mn
sin(mϕ)
]
P
m
n
(cos ϑ) (9.30)
In order to obtain a more suitable form of (9.29b) we introduce the substitutions
f (r) =
R
n
(r)
√
r
, ρ = kr (9.31)
yielding
d
2
R
n
dρ
2
+
1
ρ
dR
n
dρ
+
1 −
(n + 1/2)
2
ρ
2
R
n
= 0 (9.32)
The quantity R
n
is a new variable. Equation (9.32) is known as Bessel’s differential
equation of order (
n + 1/2).
Using the notation
R
n
(ρ) = Z
n+1/2
(ρ) (9.33)
the solution to (9.29b) is given by
f
n
(r) =
1
√
r
Z
n+1/2
(kr ) (9.34)
Employing the latter relation, the solution of the scalar wave equation in spherical
coordinates is written as
u(ρ,ϑ,ϕ,t) =
%
π
2ρ
Z
n+1/2
(ρ)F
n
(ϑ, ϕ) exp(−iωt) (9.35)
where for convenience we have included the factor
√
π/2k. This does not change the
solution at all since this factor multiplies the integration constants. By introducing
the so-called spherical Bessel functions
z
n
(ρ) =
%
π
2ρ
Z
n+1/2
(ρ) (9.36)
9.4 The solution of the wave equation 341
the elementary or characteristic wave functions can finally be written as
u
e
mn
(ρ,ϑ,ϕ,t) = cos(mϕ)P
m
n
(cos ϑ)z
n
(ρ) exp(−iωt)
u
o,mn
(ρ,ϑ,ϕ,t) = sin(mϕ)P
m
n
(cos ϑ)z
n
(ρ) exp(−iωt)
, m = 0, 1,...,n
(9.37)
Here we have distinguished between the even and odd form of the wave functions
u
e
mn
and u
o,mn
which are denoted according to the even and odd functions cos(mϕ)
and sin(mϕ) occurring in (9.30). Everywhere on the surface of the sphere the
elementary wave functions u
e
o,mn
are finite and single-valued, see also Stratton
(1941).
9.4.2 Solution of the vector wave equation in spherical coordinates
Let us reconsider the vector wave equations (9.5). Only if E and H are resolved in
terms of rectangular components we obtain three independent scalar vector wave
equations of the form (9.23) for each component of the vectors. In order to solve the
vector wave equations (9.5) in spherical coordinates, we make use of the following
two theorems.
(a) If ψ satisfies the scalar wave equation (9.23) then the vectors defined by
M
ψ
=∇×(rψ), N
ψ
=
1
k
∇×M
ψ
(9.38)
satisfy the vector wave equations
∇
2
M
ψ
+ k
2
M
ψ
= 0, ∇
2
N
ψ
+ k
2
N
ψ
= 0 (9.39)
Furthermore, the following relation is valid
M
ψ
=
1
k
∇×N
ψ
(9.40)
The quantities M and N are called vector wave functions. The proof of the above
statements will be left to the exercises. It will be observed that M and N are solenoidal
vectors, i.e. they satisfy
∇·M
ψ
= 0, ∇·N
ψ
= 0 (9.41)
(b) If u and v are two solutions of the scalar wave equation, then the vectors M
u
, M
v
,
N
u
and N
v
represent the derived vector fields which satisfy the vector wave equation.
The two vectors A and B defined by
A = M
v
− i N
u
, B =−N (M
u
+ i N
v
) (9.42)
satisfy
∇×A = ik
0
B, ∇×B =−ik
0
N
2
A (9.43)
342 Light scattering theory for spheres
In order to evaluate the boundary conditions we must know the components of
the M and N functions. Let us consider the curl of the arbitrary vector A which is
given by the well-known formula
∇×A =
1
r
2
sin ϑ
re
ϑ
r sin ϑe
ϕ
e
r
∂
∂ϑ
∂
∂ϕ
∂
∂r
rA
ϑ
r sin ϑ A
ϕ
A
r
(9.44)
Recalling the definition (9.38), we identify the vector A by ψr = ψre
r
where e
r
is
the unit vector in direction r. Thus we obtain for the components of M and N
M
ψ,ϑ
=
1
sin ϑ
∂ψ
∂ϕ
, M
ψ,ϕ
=−
∂ψ
∂ϑ
, M
ψ,r
= 0
N
ψ,ϑ
=
1
kr
∂
2
rψ
∂r ∂ϑ
, N
ψ,ϕ
=
1
kr sin ϑ
∂
2
rψ
∂r ∂ϕ
N
ψ,r
=−
1
kr sin ϑ
∂
∂ϑ
sin ϑ
∂ψ
∂ϑ
+
1
sin ϑ
∂
2
ψ
∂ϕ
2
(9.45)
The final expression in (9.45) can be stated in a much simpler form by involving
the scalar wave equation in spherical coordinates (9.24) as well as (9.29b). This
immediately leads to
N
ψ,r
= krψ +
r
k
∂
2
ψ
∂r
2
+
2
k
∂ψ
∂r
=
n(n + 1)ψ
kr
(9.46)
These relations will be needed for the evaluation of the boundary conditions.
The M and N functions obey some very desirable and helpful orthogonality
relations which will be derived now. We start the analysis by assuming a harmonic
time dependence of the two functions
M
ψ
= m
ψ
exp(−iωt), N
ψ
= n
ψ
exp(−iωt) (9.47)
With the help of (9.37) and (9.45) we obtain for the m functions
m
e
o,mn
(r,ϑ,ϕ) =
1
sin ϑ
∂ψ
e
o,mn
∂ϕ
e
ϑ
−
∂ψ
e
o,mn
∂ϑ
e
ϕ
=
− sin(mϕ)
cos(mϕ)
m
sin ϑ
P
m
n
(cos ϑ)z
n
(kr )e
ϑ
−
cos(mϕ)
sin(mϕ)
dP
m
n
dϑ
z
n
(kr )e
ϕ
(9.48)
Thus it is seen that for m = 0 the odd function m
o,m=0,n
= 0.
9.4 The solution of the wave equation 343
To find the orthogonality conditions of the M functions, we multiply m
e
o,mn
by
m
e
o,ml
. Integrating the product over the unit sphere yields
2π
0
π
0
m
e
o,mn
· m
e
o,ml
sin ϑdϑdϕ
=z
n
(kr )z
l
(kr )
2π
0
sin
2
(mϕ)
cos
2
(mϕ)
dϕ
π
0
m
2
P
m
n
(cos ϑ)P
m
l
(cos ϑ)
sin ϑ
dϑ
+z
n
(kr )z
l
(kr )
2π
0
cos
2
(mϕ)
sin
2
(mϕ)
dϕ
π
0
dP
m
n
dϑ
dP
m
l
dϑ
sin ϑdϑ (9.49)
From textbooks discussing Legendre polynomials we extract the important relation
m
2
π
0
P
m
n
(cos ϑ)P
m
l
(cos ϑ)
sin ϑ
dϑ +
π
0
dP
m
n
dϑ
dP
m
l
dϑ
sin ϑdϑ
=
2n(n + 1)
2n + 1
(n + m)!
(n − m)!
δ
n,l
(9.50)
By recalling the well-known orthogonality relations of the trigonometric functions
1
π
2π
0
cos(mx) cos(kx)dx = (1 + δ
0,m
)δ
m,k
1
π
2π
0
sin(mx) sin(kx)dx = (1 − δ
0,m
)δ
m,k
(9.51)
the orthogonality relation of the m functions can finally be stated as
2π
0
π
0
m
e
o,mn
· m
e
o,mn
sin ϑdϑdϕ
= π
(1 + δ
0,m
)
(1 − δ
0,m
)
2n(n + 1)
2n + 1
(n + m)!
(n − m)!
[
z
n
(kr )
]
2
(9.52)
Since the m functions differ from the M functions only by the time factor exp(−iωt),
we have obtained the desired result.
Utilizing (9.45) and (9.46) we obtain for the n functions
n
e
o,mn
=
1
kr
∂
2
∂r ∂ϑ
rψ
e
o,mn
e
ϑ
+
1
kr sin ϑ
∂
2
∂r ∂ϕ
rψ
e
o,mn
e
ϕ
+
n(n + 1)
kr
ψ
e
o,mn
e
r
(9.53)
344 Light scattering theory for spheres
Carrying out the differentiations we find after a few easy steps
n
e
o,mn
=
1
kr
∂
∂r
[rz
n
(kr )]
dP
m
n
dϑ
cos(mϕ)
sin(mϕ)
e
ϑ
×
−sin(mϕ)
+ cos(mϕ)
m
kr sin ϑ
∂
∂r
[rz
n
(kr )]P
m
n
(cos ϑ)e
ϕ
+
n(n + 1)
kr
z
n
(kr )P
m
n
(cos ϑ)
cos(mϕ)
sin(mϕ)
e
r
(9.54)
Again we observe that for m = 0 the odd function n
o,m=0,n
= 0.
The orthogonality relations for the n functions can be obtained analogously to
the m functions. After a few easy steps we find
2π
0
π
0
n
e
o,mn
· n
e
o,ml
sin ϑdϑdϕ
=
1
(kr )
2
∂
∂r
[rz
n
(kr )]
∂
∂r
[rz
l
(kr )]
π
0
dP
m
n
dϑ
dP
m
l
dϑ
sin ϑdϑ
2π
0
cos
2
(mϕ)
sin
2
(mϕ)
dϕ
+
m
2
(kr )
2
∂
∂r
[rz
n
(kr )]
∂
∂r
[rz
l
(kr )]
π
0
P
m
n
(cos ϑ)P
m
l
(cos ϑ)
sin ϑ
dϑ
×
2π
0
sin
2
(mϕ)
cos
2
(mϕ)
dϕ +
n(n + 1)
kr
l(l + 1)
kr
z
n
(kr )z
l
(kr )
×
π
0
P
m
n
(cos ϑ)P
m
l
(cos ϑ) sin ϑdϑ
2π
0
cos
2
(mϕ)
sin
2
(mϕ)
dϕ
(9.55)
In order to evaluate the spherical Bessel functions we make use of the following
recurrence relations
z
n−1
(r) + z
n+1
(r) =
2n + 1
r
z
n
(r),
dz
n
(r)
dr
=
1
2n + 1
[
nz
n−1
(r) −(n + 1)z
n+1
(r)
]
(9.56)