Multipole Fields II
1 Solution of Helmholtz equation in spherical coordinates
We want to ®nd exact solutions of the wave equation
1 @2Ã
=0
c2 @t2
~ E
~ ; or B
~ outside the source
where à may be the electric potential ©;or a component of A;
region. Taking the Fourier time transform, we obtain the Helmholtz equation:
¡ 2
¢
r + k2 Ã (~x ;!) = 0
r2 Ã (~x ;t) ¡
In spherical coords, we have:
µ
¶
1 @
2 @Ã
r
+ r2ang à + k2 à = 0
r 2 @r
@r
This equation is very similar to Laplacezs equation, differing only by the term in k2 : We
solve it similarly, by separating variables. The solutions are of the form:
X
Ã=
flm (r) Ylm (µ; Á)
l;m
where the functions flm satisfy the equation:
d2 f
2 df
l (l + 1)
+
+ k2 f ¡
f=0
2
dr
r dr
r2
where
Equation (1) shows that f lm
Now let
Then
and
(1)
l (l + 1)
f lm Y lm
r2
will depend on l but not on m; so henceforth we shall call it fl :
r2ang flm Y lm = ¡
u l (r)
fl (r) = p
r
df
ul
1 dul
= ¡ 3=2 + 1=2
dr
2r
r
dr
d2 f
3u l
1 du l
1 d2 u l
=
¡
+
dr2
4r 5=2
r3=2 dr
r 1=2 dr2
Thus equation (1) becomes:
µ
¶ ·
¸
3ul
1 dul
1 d2 ul
2
ul
1 du l
l (l + 1)
2
¡ 3=2
+ 1=2
+
¡ 3=2 + 1=2
+ k ¡
4r5=2
r
dr
r
dr2
r
2r
r
dr
r2
·
¸
2
1 d ul
1 du l
l (l + 1) + 1=4
2
+ 3=2
+ k ¡
r1=2 dr 2
r
dr
r2
1
u l (r)
p
r
= 0
u l (r)
p
r
= 0
Now we multiply by
p
r to get:
"
¡
¢2 #
l + 12
d2 u l
1 du l
2
+
+ k ¡
u l (r) = 0
dr2
r dr
r2
which is Besselzs equation with º = l + 1=2: Thus the solutions for fl are of the form
J l+1=2
p
r
These are the spherical Bessel functions:
r
¼ J l+1=2 (x)
jl (x) =
p
2
x
The other Bessel functions are de®ned similarly. Thus:
r
¼ Nl+1=2 (x)
nl (x) =
p
2
x
and
r
¢
¼ ¡
(1;2)
hl
(x) =
J l+1=2 (x) § iN l+1=2 (x)
2x
= j l (x) § inl (x)
The hs are particularly useful since they have wave-like behavior at large argument:
eix
for x À l
x
The jl are well behaved at both small and large arguments:
(1)
hl
(x) ' (¡i)l+1
j l (x) '
and
xl
for x ¿ 1; l
(2l + 1)!
¡
¢
sin x ¡ l ¼2
j l (x) '
for x À l
x
2 The Greenzs function
Next we expand the Greenzs function for the Helmholtz equation in these eigenfunctions.
The appropriate equation is
¡ 2
¢
r + k2 G (~x ; ~x0 ; !) = ¡± (~x ¡ ~x 0 )
Expanding G in spherical harmonics, we have:
X
¡ 0 0¢
¤
G (~x; ~x 0 ; !) =
gl (r; r0 ) Y lm (µ; Á) Ylm
µ ;Á
l;m
2
Stuf®ng in:
¸
X 1 d µ dgl ¶ ·
¡ 0 0¢
¡
¢
l (l + 1)
± (r ¡ r 0)
2
2
¤
r
+
k
¡
gl Y lm (µ; Á) Ylm
µ ; Á = ¡± (~x ¡ ~x0 ) = ¡
± (¹ ¡ ¹0 ) ± Á ¡ Á0
2
2
2
r dr
dr
r
r
l;m
¤
Now multiply both sides by Y lm
(µ; Á) and integrate over the angles. We get
µ
¶ ·
¸
¡ 0 0¢
1 d
l (l + 1)
± (r ¡ r0 ) ¤ ¡ 0 0 ¢
2 dgl
2
¤
r
+
k
¡
g
Y
µ
;
Á
=
¡
Ylm µ ; Á
l
lm
r2 dr
dr
r2
r2
so the equation for g l is:
µ
¶ ·
¸
1 d
l (l + 1)
± (r ¡ r0 )
2 dg l
2
r
+
k
¡
gl = ¡
2
2
r dr
dr
r
r2
For r < r0 we need a solution ®nite at r = 0 : the j l : We want an outgoing wave at in®nity,
( 1)
so we choose hl in the region r > r0 : Then by the symmetry of the Greenzs function, we
get:
(1)
g l (r; r0 ) = Al jl (kr< ) h l (kr> )
Next we multiply by r 2 and integrate across the boundary at r = r0 :
¯ r0 +"
¯
2 dg l ¯
r
= ¡1
dr ¯ r0 ¡"
³
´
1
(1)0
(1)
kAl hl (kr 0) jl (kr0 ) ¡ jl0 (kr 0) hl (kr 0) = ¡
2
(r 0)
The term in parentheses on the left hand side is the Wronskian of the two solutions jl and
( 1)
hl : Using the large argument form of these functions, we get:
W
l+1
=
(¡i)
=
(¡i)
=
(¡i)
=
i
x2
l+1
l+1
"
¡
¢
¡
¢
¡
¢#
ix
sin x ¡ l ¼2
cos x ¡ l ¼2
sin x ¡ l ¼2
l+1 e
¡ (¡i)
¡
x
x
x
x2
¡
¢
¡
¢ #!
³
³
eix
¼ ´ sin x ¡ l ¼2
¼ ´ sin x ¡ l ¼2
i sin x ¡ l
¡
¡ cos x ¡ l
+
x2
2
x
2
x
¡
¢
l+1
l+2 2l+1
l
l
¡il
eix ³ ¡ i(x¡l ¼2 ) ´ (¡i)
(¡1)
i
(¡1) (¡1) i
¡e
=
=
=
2
2
2
2
x
x
x
x
eix
x
µ
1
i¡
x
Ã"
¶
Thus
kAl
i
and so
Al = ¡
and thus
G (~x; ~x 0 ) =
=¡
2
(kr0 )
1
2
(r 0)
k
= ik
i
X
¡ 0 0¢
eikR
(1)
¤
= ik
jl (kr< ) h l (kr> ) Y lm (µ; Á) Ylm
µ ;Á
4¼R
l; m
3
(2)
3 Solution to the vector Helmholtz equation
The equation satis®ed by the electric and magnetic ®elds is:
¡ 2
¢
~
r + k 2 E=0
In any coordinate system other than Cartesian, we cannot separate this equation into
components because the unit vectors are not constants. Thus we need to ®nd a scalar
quantity that satis®es the Helmholtz equation. Letzs investigate
³
´
¡¡
¢
¢
~
r2 ~r ¢ E
= @i @ i r j E j = @i @ i r j E j + r j @ i E j
¡
¢
¡
¢
¡
¢
= @ i@ irj E j + @i rj @iE j + (@irj ) @ iE j + rj @i@ iE j
= 0 + ± ij @i E j + ± ij @ iE j + rj r2 E j
³
´
~ ¢E
~ + ~r ¢ r2 E
~
= 2 r
~ ¢E
~ =0; and so
But outside the source, r
³
´
~ = ~r ¢ r2 E
~
r2 ~r ¢ E
Thus we have
¡
r2 + k 2
¢³
´
¡
¢
~ = ~r ¢ r2 + k2 E
~ =0
~r ¢ E
~ is the scalar quantity we need.
and thus ~r ¢ E
~ since r
~ ¢B
~ = 0 always. The solutions for
Notice we can do the same thing with B;
~ and ~r ¢ B
~ will thus be of the form:
~r ¢ E
~ ¡ l (l + 1) gl (kr) Y lm (µ; Á)
~r ¢ E=
k
l(l+1)
where the constant ¡ k has been inserted for future convenience, and the function gl
must be a combination of the spherical Bessel functions, for example
(1) (1)
(2) (2)
gl = Al h l (kr) + Al h l
(kr)
~ will be given by:
Then the corresponding B
~ = ¡ir
~ £E
~ = 1r
~ £E
~
B
k
ik
We can borrow some ideas from quantum mechanics. The angular momentum operator
is:
~ = 1 ~r £ r
~
L
i
and its eigenfunctions are the Ylm : Notice that
~
~r ¢ L=0
Then
³
´
³
´
~ = 1 ~r ¢ r
~ £E
~ = 1 ~r £ r
~ ¢E
~ = ~L ¢ E
~
k~r ¢ B
i
i
Thus we may de®ne the following eigenfunctions for the problem:
4
(3)
~ purely tangential:
The electric or transverse magnetic mode has B
~ (E) = ¡ l (l + 1) gl (kr) Y lm (µ; Á)
~r ¢ E
lm
k
~ (E) = 0
~r ¢ B
lm
~ purely tangential:
The magnetic or transverse electric mode has E
~ (M ) = l (l + 1) gl (kr) Ylm (µ; Á)
~r ¢ B
lm
k
(M )
~
~r ¢ E
= 0
lm
Thus from equation (3),
~ ¢E
~ (M ) = l (l + 1) gl (kr) Y lm (µ; Á)
L
lm
(4)
~
4 Properties of L
1
~
~r £ r
i
1
=
f^
x (y@z ¡ z@y ) + ^
y (z@x ¡ x@z ) + ^
z (x@y ¡ y@x )g
i
Now we want to write everything in spherical coordinates while keeping the Cartesian
components:
~L
=
~ cartesian
r
Thus
iLx
iLy
=
@x isph
@jspherical
j
@xcart
0
sin µ cos Á cos µrcos Á
= @ sin µ sin Á cos µrsin Á
cos µ
¡ sinr µ
Á
¡ rsin
sin µ
cos Á
r sin µ
0
1
Ar
~ spher ical
µ
¶
µ
¶
sin µ
cos µ sin Á
cos Á
r sin µ sin Á cos µ@r ¡
@µ ¡ r cos µ sin µ sin Á@r +
@µ +
@Á
r
r
r sin µ
= ¡ sin Á@µ ¡ cos Á cot µ@Á
µ
¶
µ
¶
cos µ cos Á
sin Á
sin µ
= r cos µ sin µ cos Á@r ¡
@µ ¡
@Á ¡ r sin µ cos Á cos µ@r ¡
@µ
r
r sin µ
r
= cos Á@µ ¡ cot µ sin Á@Á
µ
¶
cos µ sin Á
cos Á
iLz = r sin µ cos Á sin µ sin Á@r +
@µ +
@Á
r
r sin µ
µ
¶
cos µ cos Á
sin Á
¡r sin µ sin Á sin µ cos Á@r +
@µ ¡
@Á = @Á
r
r sin µ
=
5
Lz is very nice. We can form linear combinations of Lx and Ly that look almost as nice:
L§
=
=
=
=
Lx § iLy
1
(¡ sin Á@µ ¡ cos Á cot µ@Á § i (cos Á@µ ¡ cot µ sin Á@Á ))
i
(§ cos Á@µ + i sin Á@µ + i cos Á cot µ@Á ¨ cot µ sin Á@Á )
e§iÁ (§@µ + i cot µ@Á )
Then
~ ¢L
~
L
=
L2x + L2y + L2z
=
¡ (sin Á@µ + cos Á cot µ@Á )2 ¡ (cos Á@µ ¡ cot µ sin Á@Á )2 ¡ @2Á
=
¡@µ2 ¡ sin Á cos Á@µ cot µ@Á ¡ cos2 Á cot µ@µ ¡ cos2 Á cot2 µ@2Á
+ cos Á sin Á@µ cot µ@Á ¡ cot µ sin Á2 @µ ¡ cot2 µ sin 2 Á@Á2 ¡ @Á2
¡
¢
= ¡@µ2 ¡ cot µ@µ ¡ cot2 µ + 1 @Á2
1
1
= ¡
@µ (sin µ@µ ) ¡
@2
sin µ
sin 2 µ Á
which is the angular part of r2 : Thus:
(5)
L 2Y lm = l (l + 1) Y lm
We also have
Lz Ylm = mY lm
~
5 Complete solution for E:
Suppose
~ = Lg
~ l (kr) Y lm (µ; Á)
E
Then
~L ¢ E
~ = gl L2 Y lm = l (l + 1) gl Y lm
which is exactly what we need (cf equation 4). Thus the solutions are:
Magnetic or transverse electric modes:
~ (M )
E
lm
=
~ (M )
B
lm
=
gl (kr) ~LYlm (µ; Á)
i~
~ (M )
¡ r
£E
lm
k
(6)
and
Electric or transverse magnetic modes:
~ (E )
B
lm
~ (E )
E
lm
~ lm (µ; Á)
= g l (kr) LY
i~
~ (E)
=
r£B
lm
k
6
(7)
Now we de®ne the vector spherical harmonics:
~ lm = p 1
~ lm
X
LY
l (l + 1)
so that
~ (M ) = gl (kr) X
~ lm
E
lm
which are the appropriate eigenfunctions for the problem.
Combining the modes we get the general solution:
¸
X·
~ =
~ lm + i aE (l:m) r
~ £ f l (kr) X
~ lm
E
aM (l; m) gl (kr) X
k
l;m
·
¸
X
~ =
~ lm ¡ i aM (l:m) r
~ £ g l (kr) X
~ lm
B
aE (l; m) fl (kr) X
k
l;m
(8)
6 Finding the coef®cients
~ lm = 0;then
Note that, since ~r ¢ X
³
´
X i
~ =
~ £ g l (kr) X
~ lm
~r ¢ B
¡ aM (l:m) ~r ¢ r
k
l;m
³
³
´
³
´
´
X
1
~ l £ ~r £ r
~ Y lm + g l r£
~
~ Y lm
=
¡ p
aM (l:m) ~r ¢ rg
~r £ r
k l (l + 1)
l;m
³
´
~ l is in the ^r direction, and so rg
~ l £ ~r £ r
~ Y lm is perpendicular to ^
But rg
r; and so its dot
product with ~r is zero. Also:
~ £ ~L
ir
=
=
and
Thus: Thus:
³
´
¡
¢
¡
¢
~ £ r~ £ r
~ = @j r i @ j ¡ @ j r j @ i
r
±ij @ j + r ir2 ¡ 3@ i ¡ rj @ i@ j = ¡ri r2 ¡ 2@i ¡ rj @i @ j
³
´
~ 1 + ~r ¢ r
~
r
¡
¢
= @ i 1 + rj @ j = @i + ±ij @ j + rj @ i@ j
= 2@ i + rj @ i@ j
³
´
~ £ L=~
~ r r2 ¡ r
~ 1 + ~r ¢ r
~
ir
7
(9)
~
~r ¢ B
³
³
´
´
1
~ £ ~r £ r
~ Ylm
¡ p
aM (l:m) gl (kr) ~r ¢ r
k l (l + 1)
l;m
³
³
´
´
X
1
~ 1 + ~r ¢ r
~ Y lm
=
¡ p
aM (l:m) gl (kr) ~r ¢ ~rr2 ¡ r
k l (l + 1)
l;m
³
³
´
´
X
1
~ 1 + ~r ¢ r
~ Y lm
=
¡ p
aM (l:m) gl (kr) r2 r2 ¡ ~r ¢ r
k l (l + 1)
l;m
=
X
~ lm = 0; and we already know r2 Y lm :
But Y lm is independent of r; so ~r ¢ rY
X
1
¡l (l + 1)
~ =
~r ¢ B
¡ p
aM (l:m) g l (kr) r 2
Y lm
r2
k l (l + 1)
l;m
p
X l (l + 1)
=
aM (l:m) gl (kr) Y lm
k
(10)
l;m
Then using the orthogonality of the Y lm ;we have:
Z
k
¤
~ lm
aM (l:m) g l (kr) = p
~r ¢ BY
dl (l + 1)
Similarly we ®nd:
Z
¡k
~ Y ¤ daE (l:m) fl (kr) = p
~r ¢ E
lm
l (l + 1)
7 Properties of the ®elds
7.1
The near zone
For kr ¿ 1; we may use the small argument expansion of the Bessel functions:
l
j l » (kr)
while
(2l ¡ 1)!!
nl » ¡
xl+1
Thus both hs behave like ns for kr ¿ 1: Then for the electric mode we have:
~ (E) » ~L Y lm (µ; Á)
B
lm
rl+1
and
~ (E) = ¡ i r
~ £B
~ (E) = ¡ i r
~ £L
~ Y lm (µ; Á)
E
lm
lm
k
k
r l+1
and using equation (9) we may write the electric ®eld as:
h
³
´i Y (µ; Á)
lm
~ (E) = ¡ 1 ~r r2 ¡ r
~ 1 + ~r ¢ r
~
E
lm
k
r l+1
8
(11)
The ®rst term is zero because
Ylm (µ; Á)
rl+1
~ (E )
E
lm
Y
is a solution of Laplacezs equation. Thus:
´
1~ ³
~ Y lm (µ; Á)
=
r 1 + ~r ¢ r
k
rl+1
1~
Ylm (µ; Á)
=
r (1 ¡ (l + 1))
k
rl+1
l ~ Y lm (µ; Á)
= ¡ r
k
rl+1
(µ ;Á)
Now lm
is a solution of Laplacezs equation and thus represents an electrostatic
rl+1
potential: it is a static multipole potential, and its gradient therefore represents a static
multipole ®eld. Thus in the near zone the electric mode is a static electric multipole ®eld
~ (k)): This is the origin
that oscillates in time (because of the e¡ i!t factor that multiplies E
of the name }electric} for this mode. Similarly, the magnetic ®eld in the magnetic mode has
a static multipole character in the near zone.
7.2
The far zone
When kr À 1 we may expand the Bessel functions using the large argument expansion.
(1)
l+1
h l (kr) » (¡i)
and so
eikr
kr
ikr
~ (E) » (¡i)l+1 e ~LY lm
B
lm
kr
and
~ (E)
E
lm
l+1
»
=
=
ikr
(¡i)
~ e LY
~ lm
r£
k
kr
µ
¶
(¡i)l+ 2 ~ e ikr ~
eikr ~
~
¡
r
£LY lm +
r £ LYlm
k2
r
r
µ·
¸
¶
h
³
´i
(¡i) l eikr
^r
2
~
~ lm + 1 ~
~ 1 + ~r ¢ r
~
i
k¡
£
LY
r
r
¡
r
Y
lm
k2
r
r
i
i
~ lm ´ 0: and so dropping terms of order 1=kr; we
where we used equation (9). Again ~r ¢ rY
have:
µ
·
¸
¶
l
(¡i) eikr
1
¡l (l + 1)
(E)
~
^
~
~
Elm =
i k £ LY lm +
~r
¡ r Ylm
k
r
ik
r2
Both terms in the square brackets are of order 1=kr copmared with the ®rst term, and so we
drop them. Then:
~ ( E)
E
lm
and the ®eld is a radiation ®eld.
eikr ^ ~
k £ LYlm
kr
=
¡ (¡i)l+1
=
)
~ (E
¡^
k£B
lm
9
(12)
~ lm
8 Orthogonality of the X
~ lm are orthogonal, that is
Our next task is to demonstrate that the X
Z
~ lm ¢ X
~ ¤l 0m 0 d- = ± ll 0 ±mm0
X
(13)
sphere
~ lm is terms of the Y lm :
The ®rst step is to evaluate X
µ ·
¸
·
¸
¶
1
1
L+ + L¡
L+ ¡ L¡
~
~
Xlm = p
LY lm = p
x
^
+y
^
+^
zLz Y lm
2
2i
l (l + 1)
l (l + 1)
So we need to evaluate L+ Ylm and L¡ Ylm :
L§ Ylm
=
=
=
e§iÁ (§@µ + i cot µ@Á ) Y lm
s
2l + 1 (l ¡ m)! § iÁ
e
(§@µ + i cot µ@Á ) P lm (cos µ) eimÁ
4¼ (l + m)!
s
2l + 1 (l ¡ m)!
(§@µ ¡ m cot µ) P lm (cos µ) ei(m§ 1)Á
4¼ (l + m)!
Now
¹ ´ m
P l (¹)
sin µ
¢
¢
1 ¡ ¡
=
¨ 1 ¡ ¹2 P lm0 (¹) ¡ m¹P lm (¹)
sin µ
But from the m-raising and m¡lowering recursion relations for the Legendre functions (eg
Problem 8.16 pg 431 in Lea1 ),
p
¡
¢
1 ¡ ¹2 Plm0 (¹) = ¡m¹Plm ¡ 1 ¡ ¹2 Plm+ 1
p
= m¹P lm + (l + m) (l ¡ m + 1) 1 ¡ ¹2 Plm¡ 1
(§@µ ¡ m cot µ) P lm (cos µ) =
³
¨ sin µ@¹ ¡ m
and thus:
(+@µ ¡ m cot µ) P lm (cos µ) =
=
=
and so
2l + 1 (l ¡ m)! m+1 i(m+1)Á
P
e
4¼ (l + m)! l
s
(l ¡ m)! (l + m + 1)!
=
Y l;m+1
(l + m)! (l ¡ m ¡ 1)!
L+ Y lm =
L+ Y lm
s
¢
¢
1 ¡ ¡
¡ 1 ¡ ¹2 P lm0 (¹) ¡ m¹P lm (¹)
sin µ
´
1 ³p
1 ¡ ¹2 P lm+1 + m¹P lm ¡ m¹P lm (¹)
sin µ
1 p
1 ¡ ¹2 P lm+1 = Plm+1
sin µ
1
Note: there is some disagreement in the literature as to the de®nition of Plm: My choice is the same as
Jacksonzs. Some authors (eg Butkov) have a de®nition that differs by a factor (¡1)m :
10
L+ Y lm =
Similarly
(¡@µ ¡ m cot µ) P lm (cos µ) =
=
=
and thus
L¡ Ylm
s
=
s
p
(l ¡ m) (l + m + 1)Y l;m+1
¢
¢
1 ¡ ¡
+ 1 ¡ ¹2 P lm0 (¹) ¡ m¹P lm (¹)
sin µ
´
p
1 ³
m¹Plm + (l + m) (l ¡ m + 1) 1 ¡ ¹2 Plm¡ 1 ¡ m¹Plm (¹)
sin µ
(l + m) (l ¡ m + 1) P lm¡1
2l + 1 (l ¡ m)!
(l + m) (l ¡ m + 1) Plm¡1 ei(m¡ 1)Á
4¼ (l + m)!
2l + 1
(l ¡ m + 1)!
(l + m) (l ¡ m + 1) Plm¡1 ei(m¡ 1)Á
4¼ (l ¡ m + 1) (l + m ¡ 1)! (l + m)
p
(l + m) (l ¡ m + 1)
p
Y l; m¡1 = (l + m) (l ¡ m + 1)Y l; m¡1
(l ¡ m + 1) (l + m)
=
=
The quantity in the square root may be put in a nicer form:
(l + m) (l ¡ m + 1) = l 2 + l ¡ m 2 + m = l (l + 1) ¡ m (m ¡ 1)
Thus
L¡ Ylm =
and similarly
L+ Ylm =
Then:
~ lm
X
=
=
Now note that:
p
l (l + 1) ¡ m (m ¡ 1)Yl;m¡ 1
p
l (l + 1) ¡ m (m + 1)Yl;m+ 1
µ ·
¸
·
¸
¶
L+ + L¡
L+ ¡ L¡
p
x
^
+^
y
+^
zLz Y lm
2
2i
l (l + 1)
p
µ
¶
1
1
l (l + 1) ¡ m (m + 1)Yl;m+1 (^
x¡i^
y)
2
p
p
(14)
+ 12 l (l + 1) ¡ m (m ¡ 1)Yl;m¡ 1 (^
x + i^
y) + mY lm ^z
l (l + 1)
1
µ
¶ µ
¶¤
x¡i^
^
y
x ¢^
^
x¡^
y ¢y
^
¢
=
=0
2
4
and similarly
µ
¶ µ
¶¤
x+i^
^
y
x
^ +i^
y
1
¢
=
2
2
2
Then the orthogonality integral becomes:
Z
~ lm ¢ X
~ ¤0 0 dX
l m
x+i^
^
y
2
sphere
=
p
2 p
3
¤
0 0
0
0
l
(l
+
1)
¡
m
(m
+
1)
p
pl (l + 1) ¡ m (m + 1)Yl;m+ 1 Yl 0;m 0+ 1
1
4 + l (l + 1) ¡ m (m ¡ 1) l0 (l 0 + 1) ¡ m0 (m 0 ¡ 1)Yl;m¡ 1 Y ¤0 0
5 dl ; m ¡1
2l (l + 1) sph
0
¤0
+2mm Y lm Ylm
Z
11
Z
£
¤
1
l (l + 1) ¡ m (m + 1) + l (l + 1) ¡ m (m ¡ 1) + 2m2 ±ll0 ±mm 0
2l (l + 1)
sphere
1
=
2l (l + 1) ±ll0 ± mm 0 = ±ll 0 ±mm0
2l (l + 1)
where we used the orthogonality of the Y lm :
Also
µ
¶ µ
·
¸¶
Z
Z
³
´
1
1
~ ¤0 0 ¢ ~r £ X
~ lm d- =
~ ¤0 0 ¢ ~r £ 1 ~r £ rY
~ lm
X
~
r
£
rY
dl m
l m
l (l + 1) sphere i
i
sphere
~ lm ¢ X
~ ¤0 0 dX
l m
=
h
i
³
´
~ lm = ~r ~r ¢ rY
~ lm ¡ r2 rY
~ lm = ¡r 2 rY
~ lm
~r £ r~ £ rY
where
Thus:Z
spher e
Now
³
´
~ ¤0 0 ¢ ~r £ X
~ lm d- =
X
l m
~ £ ~r=0;so:
and r
Z
~ ¤0
X
sphe re
l m0
¡1
l (l + 1)
Z
sphere
³
´ ³
´
~ ¤0 0 ¢ ¡r2 rY
~ lm d~r £ rY
l m
~ £ (~r Y ¤0 0 ) = rY
~ ¤0 0 £ ~r + Y ¤0 0 r
~ £ ~r
r
l m
l m
l m
³
´
~ lm d- =
¢ ~r £ X
r2
l (l + 1)
Z
sphere
³
´
~ £ (~rY ¤0 0 ) ¢ rY
~ lm d¡r
l m
³
´
³
´
³
´
~ ¢ ~a £ ~b = ~b ¢ r
~ £ ~a ¡ ~a ¢ r
~ £ ~b
r
~ lm and ~b = ~rY ¤0 0 ; we have r
~ £ ~a = 0;and then
so taking ~a = rY
l m
Z
Z
³
´
³
´
2
r
~ l¤0 m0 ¢ ~r £ X
~ lm d- =
~ ¢ rY
~ lm £ ~r Y l¤0 m0 dX
r
l (l + 1) sphere
sphere
And
Then
Z r2 Z
r1
sphe re
³
´
~ ¤0 0 ¢ ~r £ X
~ lm d-dr
X
l m
=
=
Z
³
´
1
~ ¢ rY
~ lm £ ~rY ¤0 0 dV
r
l m
l (l + 1) spher ical shell
Z
³
´
1
~ lm £ ~r Yl¤0m 0 dA
^r¢ rY
l (l + 1) spher ical shell
~ lm is tangential, and therefore so is rY
~ lm £ ~r; and
Here the integrand is zero, because rY
thus its dot product with ^r is zero. Thus the volume integral is zero for arbitrary values of r1
and r2 : So we must conclude that
Z
³
´
~ ¤0 0 ¢ ~r £ X
~ lm d- = 0
X
(15)
l m
sphe re
Finally
Z
Z
³
´ ³
´
~ lm ¢ ~r £ X
~ ¤0 0 d- =
~r £ X
l m
sphere
sphere
h
³
´³
´i
~ lm ¢ X
~ ¤0 0 ¡ ~r ¢ X
~ lm ~r ¢ X
~ ¤0 0 dr2 X
l m
l m
(16)
12
The last term is identically zero, and so:
Z
³
´ ³
´
~ lm ¢ ~r £ X
~ l¤0m 0 d- = r 2 ±ll 0 ±mm0
~r £ X
(17)
sphere
9 Energy and power
The time-averaged energy density in the ®elds is
1 ³ ~ ~ ¤ ~ ~ ¤´
u=
E¢E +B¢B
16¼
and so the energy contained in a spherical shell of thickness dr is:
Z
dU = r2 dr ud¸
Z
X·
r2 dr
~ lm + i aE (l; m) r£f
~
~
=
daM (l; m) gl (kr) X
l (kr) X lm ¢
16¼
k
l;m
·
¸¤
X
i
0
0
0
0 ~
~
~
0
0
0
0
0
0
aM (l ; m ) g l (kr) Xl m + aE (l ; m ) r£fl (kr) X l m
+ (B terms)
k
0
0
l ;m
To simplify, note that
~
~
r£f
l (kr) X lm
=
=
=
@f
~ lm + f r
~ £X
~ lm
r£ X
^
@r
h
³
´i
@f
~ lm + p f
~ 1 + ~r ¢ r
~
r£ X
^
r r2 ¡ r
~
Ylm
@r
i l (l + 1)
³
´
@f
~ lm + p f
~ Y lm
r£ X
^
~rr2 ¡ r
@r
i l (l + 1)
In the far zone, kr À 1; @f
@r » ikf while the terms in gradients of Ylm are of order f =r and
are therefore smaller by a factor of 1=kr: We ignore these terms. Then:
Z
¸
X·
r 2 dr
~ lm + i aE (l; m) @ f l (kr) ^
~ lm ¢
dU =
daM (l; m) gl (kr) X
r £X
16¼
k
@r
l;m
·
¸¤
X
i
0
0
0
0 @
~
~
aM (l ; m ) gl 0 (kr) Xl 0m 0 + aE (l ; m )
fl 0 (kr) ^r £ X l0 m0
k
@r
0
0
l ;m
and we make use of the orthogonality integrals to get:
¯
¯2
¯i
¯
r 2dr X
@
dU =
jaM (l; m) gl (kr)j2 + ¯¯ aE (l; m) f l (kr)¯¯ + (B terms)
16¼
k
@r
l;m
Now in the radiation zone,
and so
gl » f l » (¡i)
13
l+1
eikr
kr
jgl j »
and
1
kr
@f
» ikf
@r
Thus the energy is:
dU =
dr X
2
2
jaM (l; m)j + jaE (l; m)j
8¼k 2 l; m
where the change from the factor 16 to 8 takes into account the equal contribution from the
magnetic ®eld.
The total time-averaged power radiated is
dU
c X
2
2
P =c
=
jaM (l; m)j + jaE (l; m)j
(18)
dr
8¼k2
l;m
To ®nd the angular distribution of power radiated, we start with:
¯
¯
¯
dP
c 2 ¯¯ ~
~ ¤ ¯¯ = c r2 ¯¯E
~ ¢E
~ ¤ ¯¯
<
>=
r ¯E £ B
d- ¯ 8¼
h 8¼
i
¯ P
l+1
~ lm ¡ a E (l; m)^r £ X
~ lm ¢
(¡i)
aM (l; m) X
c ¯¯
l;m
h
i¤
=
P
l0 +1
0
0 ~ 0 0
0
0
8¼k2 ¯¯
~ l0 m0
i
a
(l
;
m
)
X
¡
a
(l
;
m
)
r
^
£
X
0
0
M
l
m
E
l ;m
¯
¯
¯
¯
¯
¯
(19)
where we have used the large argument approximation to the h (1) : Note we must include
the factor i l+1 to be sure we add the different spherical harmonics correctly. The electric
and magnetic multipoles for a given l; m have the same angular dependence but different
polarizations.
For a single multipole of order (l; m) we have:
i h
i¯
dP (l; m)
c ¯¯h
~ lm ¡ aE (l; m) ^r £ X
~ lm ¢ aM (l; m) X
~ lm ¡ aE (l; m) ^r £ X
~ lm ¯¯
<
>=
¯ aM (l; m) X
2
d-µ
8¼k
¯
¯2
¯
¯2 ¶
c
¯
¯
2¯~
2¯
~
=
jaM (l; m)j ¯ Xlm ¯ + jaE (l; m)j ¯^r £ X lm ¯
2
8¼k
¯2
´¯
c ³
¯
2
2 ¯~
=
ja
(l;
m)j
+
ja
(l;
m)j
X
¯
¯
M
E
lm
8¼k2
¯
¯2 ¯
¯2
¯
~ lm ¯¯ = ¯¯X
~ lm ¯¯ (see equation 16).
where we used the fact that ¯^
r £X
¯
¯
¯ ~ ¯2
We can use equation (14) to write ¯X
lm ¯ in terms of the angles.
µ
¶
¯
¯
1
l (l + 1) ¡ m (m + 1)
l (l + 1) ¡ m (m ¡ 1)
¯ ~ ¯2
2
2
2
2
¯X lm ¯
=
jY l;m+1 j +
jY l;m¡1 j + m jYlm j
l (l + 1)
2
2
Ã
!
¡ m+1 ¢2
l(l+1)¡m(m+1) (l¡ m¡1)!
Pl
+
2l + 1
2
(l+m+1)!
=
l(l+1)¡m( m¡1) (l¡m+1)! ¡ m¡1 ¢2
4¼l (l + 1)
Pl
+ m 2 ( l¡m)! (Plm )2
2
(l+m¡1)!
The ®rst few values are:
Dipole: l = 1
14
( l+m)!
m=0
¯
¯
3 ¡ 1 ¢2
3
¯ ~ ¯2
P1 =
sin 2 µ
¯X 10 ¯ =
8¼
8¼
1
0.5
0
-0.5
-1
m = §1
¯
¯
¯ ~ ¯2
¯X11 ¯
¶2
2
1 ¡ 1¢2
2
(P 1 ) +
P1
2
2
·
¸
·
¸
3
sin 2 µ
3
1 ¡ cos2 µ
cos2 µ +
=
cos2 µ +
8¼
2
8¼
2
¡
¢
3
cos 2 µ + 1
16¼
3
8¼
=
=
=
µ
0.4
0.2
-1
0
-0.8 -0.6 -0.4 -0.2
-0.2
-0.4
Quadrupole: l = 2
15
0.2
0.4
0.6
0.8
1
m=0
¯
¯
¯ ~ ¯2
¯X 20 ¯
=
=
5 6 ¡ 1 ¢2
5
2
P
=
( ¡3 sin µ cos µ)
24¼ 3! 2
24¼
15
15
sin 2 µ cos2 µ =
sin2 2µ
8¼
32¼
0.8
0.6
0.4
0.2
0
-0.8 -0.6 -0.4 -0.2
-0.2
0.2 0.4 0.6 0.8
-0.4
-0.6
-0.8
m=1
¯
¯
¯ ~ ¯2
¯ X21 ¯
=
=
µ
¶
¢2 15
1 6 ¡ 2 15
6 5 ¡
sin 4 µ +
3 cos2 µ ¡ 1 +
sin2 µ cos2 µ
6
2 32¼
2 16¼
8¼
¢
5 ¡
4 cos4 µ ¡ 3 cos2 µ + 1
16¼
0.5
-1.5
-1
-0.5
0
0.5
1
1.5
-0.5
m=2
¯
¯
¯ ~ ¯2
¯ X22 ¯
=
=
1
6
µ
6 ¡ 2 15
1 15
sin 2 µ cos 2 µ + 4
sin4 µ
2 8¼
16 2¼
¢
5 ¡
1 ¡ cos4 µ
16¼
16
¶
0.8
0.6
0.4
0.2
0
-0.4 -0.2
-0.2
0.2 0.4
-0.4
-0.6
-0.8
If we have a source with aM (l) independent of m then
¯
¯
¯
h
i ¤ ¯¯
X
0
dP
c ¯¯X
l+1
~ lm ¢
~ l0 m0 ¯
<
>=
(¡i) a M (l) X
i l +1 aM (l0 ) X
¯
d8¼k2 ¯¯
¯
l;m
l0 ;m 0
¯
¯
¯X
¯
X
¯
c ¯¯
l+l0
¤
~ lm ¢ X
~ ¤l 0m 0 ¯
=
(¡i)
aM (l) a M (l0 )
X
2
¯
¯
8¼k ¯ 0
¯
l;l
m;m0
If the modes superpose incoherently, the expression simpli®es:
¯
¯2
+l
¯
X
dP
c ¯¯X
l+1
~ lm ¯¯
<
>=
(¡i)
a
(l)
X
¯
M
d8¼ k2 ¯
¯
l
X
m=¡l
+l ¯
¯
X
¯ ~ ¯2
¯ Xlm ¯
=
c
8¼k2
=
c X
2 2l + 1
jaM (l)j
8¼k2
4¼
l
2
jaM (l)j
m=¡ l
l
(cf equation 3.69) and the radiation is isotropic. This is usually the case in atomic systems.
The total power radiated is given by the integral of (19) over solid angle. Because of the
~ lm; the interference terms do not contribute and we regain equation
orhogonality of the X
(18).
10 Angular momentum
The angular momentum density is
m
~ = ~r £ ~
p=
1 ³ ³ ~ ~ ´´
~r£ E £ B
4¼c
17
and taking the time average, we get
³ ³
´´
1
~ £B
~¤
Re ~r£ E
8¼c
³ ³
´
³
´´
1
~ ~r ¢ B
~¤ ¡B
~ ¤ ~r ¢ E
~
=
Re E
8¼c
³ ³
´¤
³
´´
1
~ L
~¢E
~ +B
~¤ L
~ ¢B
~
=
Re E
8¼kc
~ For the electric modes the ®rst term
where we used equation (3) and its equivalent for ~
r ¢ E:
is zero and for the magnetic modes the second term is zero.
Letzs investigate this expression for the magnetic modes. Using equations (10) and (8),
we get
0
1¤
p
i
0
0
X
Xh
1
l (l + 1)
~ lm
@
<m
~ >=
aM (l0 ; m 0) gl0 (kr) Y l0 m0 A
aM (l; m) g l (kr) X
8¼kc 0 0
k
<
m
~ >=
l ;m
l; m
and the total angular momentum in a spherical shell between r and r + dr is:
p
Z ³
´¤
X X l0 (l0 + 1)
r2 dr
~
~ lm Yl 0m 0 ddM =
Re
p
a ¤M (l0 ; m0 ) g ¤l0 (kr) gl aM (l; m)
LY
8¼kc
k l (l + 1)
l0 ;m0 l;m
~ lm in equation (14)
Now we use the expression we derived for LY
0
³
´
1¤
p
Z ³
Z
´¤
l (l + 1) ¡ m (m + 1)Yl;m+1 ^x¡2i^y +
~ lm Y l0 m0 d- = ¡ @ p
³
´
A Y l0 m 0 dLY
y
l (l + 1) ¡ m (m ¡ 1)Y l;m¡1 ^x+i^
+ mYlm ^
z
2
Thus we get contributions only for l = l0 and m0 = m ¡ 1; m; or m + 1:
2
³
´
3
p
Z ³
y
´¤
0+
l (l + 1) ¡ m (m + 1) ^x+i^
±
m+1;m
~
³
´2
5
LY lm Y l0 m 0 d- = ±ll0 4 p
y
l (l + 1) ¡ m (m ¡ 1) ^x¡2i^
±m¡1;m 0 + m±mm0
and thus
~ =
dM
2
r dr
Re
8¼kc
X
l;m
aM (l; m)¤ aM
³
´
3
x+i^
^
y
l
(l
+
1)
¡
m
(m
+
1)
±
0+
m+1;
m
³
´2
5
(l; m 0 ) jfl (kr)j2 4 p
y
l (l + 1) ¡ m (m ¡ 1) ^x¡2i^
±m¡ 1;m 0 + m±mm0
In the radiation zone, jf l (kr)j2 =
2
1
(kr )2
p
µ p
¶
pl (l + 1) ¡ m (m + 1)aM (l; m + 1)
+ l (l + 1) ¡ m (m ¡ 1)aM (l; m ¡ 1)
l;m
µ p
¶
X
dr
l (l + 1) ¡ m (m + 1)aM (l; m + 1)
¤
p
dMy =
Im
a
(l;
m)
M
16¼k3 c
¡ l (l + 1) ¡ m (m ¡ 1)aM (l; m ¡ 1)
l;m
dr X
2
dMz =
m jaM (l; m)j
8¼k3 c
X
dr
¤
dMx =
Re
aM (l; m)
16¼k 3c
l;m
18
and comparing with equation (18), we ®nd that for a single multipole
dM z
m dU
=
dr
! dr
suggesting that the radiation from an (l; m) multipole carries off mh units of z¡component
of angular momentum per photon of energy h!: This is consistent with the quantum
mechanical interpretation. However the x and y¡components are more complicated. See
Jackson for a more extensive discussion.
11 Connection with sources
We have already noted that the coef®cients aE and aM are determined by the radial
~ and B
~ evaluated on some spherical surface. Now we want to relate these
components of E
quantities to the properties of the sources - the charges and currents. (See Jackson for the
additional sources of magnetization.)
As usual we Fourier transform all physical quantities:
Z
1
½ (~
x; t) = p
½ (~
x;!) e¡ i!t d!
2¼
Now we de®ne a new ®eld-like quantity
~ = E+
~ 4¼i ~
F
J
!
so that
¹ ¢F
~=r
~ ¢ E+
~ 4¼i r
~ ¢J
~=r
~ ¢ E+
~ 4¼i (i!½) = r
~ ¢E
~ ¡ 4¼½ = 0
r
!
!
~ ¢J
~ with i!½: Outside the source, F
~ = E:
~
where we used charge continuity to replace r
Now
¹ £F
~ =r
~ £ E+
~ 4¼i r
~ £J
~ =ikB+
~ 4¼i r
~ £~
r
J
!
!
and
µ
¶
!~
~ £B
~ = 4¼ ~
~ 4¼i ~
~
r
J¡i E
= ¡ik E+
J = ¡ik F
c
c
!
Now we ®nd the wave equation for ~
F in the usual way:
µ
¶
³
´
4¼ i ~
~
~
~
~
~
~
r£ r £ F
= r£ ikB+
r£ J
!
³
´
³
´
³
´
~ r
~ ¢F
~ ¡ r2 F
~ = ik ¡ikF
~ + 4¼i r£
~
~ £J
~
r
r
!
So we have
´
¡ 2
¢
4¼i ~ ³ ~
r + k2 ~
F=¡
r£ r £ ~
J
!
19
and similarly
³
´
~
~ £B
~
r£
r
³
´
~ r
~ ¢B
~ ¡ r 2B
~
r
~ £F
~
= ¡ik r
µ
¶
4¼i ~ ~
~
= ¡ik ikB+
r£J
!
¡ 2
¢
~ = ¡ 4¼ r
~ £J
~
r + k2 B
c
~ and ~
~ as we did in section 3.
Now we convert these equations to equations in ~r ¢ F
r ¢B
³
´
¡ 2
¢
~ = ¡ 4¼i ~r ¢ r£
~
~ £~
r + k 2 ~r ¢ F
r
J
!
4¼ ~ ³ ~ ~ ´
=
L¢ r £ J
!
and
¡ 2
¢
4¼i ~ ~
~ = ¡ 4¼ ~r ¢ r
~ £~
r + k2 ~r ¢ B
J= ¡
L¢J
c
c
(cf equation 3). Now we can use the Greenzs function that we found previously (eqn. 2):
Z ikR
³ 0
´
e
~ = ¡1
~0 ¢ r
~ £~
~r ¢ F
L
J (~
x0 ) d 3 ~
x0
!
R
Z X
´
¡ 0 0¢ 0 ³ 0
¡ik
(1)
¤
~¢ r
~ £~
=
4¼
jl (kr< ) h l (kr> ) Y lm (µ; Á) Ylm
µ ;Á L
J (~
x0 ) d 3 ~
x0
!
l; m
and
~
r¢B
~
=
=
Z ikR
i
e
~ 0 ¢~
L
J (~
x0 ) d 3 ~
x0
c
R
Z X
¡ 0 0¢ 0
¡k
(1)
¤
~ ¢~
4¼
jl (kr< ) h l (kr> ) Y lm (µ; Á) Ylm
µ ;Á L
J (~
x0 ) d 3 ~
x0
c
l;m
then equation 11 for the coef®cient becomes:
Z
Z X
¡ 0 0¢ 0
k4¼
¡k
( 1)
0
0
¤
~ ¢J
~ (~
aM (l ; m ) gl0 (kr) = p
j l (kr< ) hl (kr> ) Ylm (µ; Á) Y lm
µ ;Á L
x 0) d 3 ~
x0 Yl¤0m 0 (µ
0
0
c
l (l + 1)
l;m
Z
¡ 0 0¢ 0
¡4¼k2
( 1)
¤
~ ¢~
aM (l; m) gl (kr) =
p
hl (kr) jl (kr0 ) Y lm
µ ;Á L
J (~
x0 ) d 3 ~
x0
c l (l + 1)
where in the last step we took r> = r outside the source.
~ ´E
~ outside the source, so:
Similarly for aE we use the fact that F
Z
Z X
´
¡ 0 0¢ 0 ³ 0
¡k4¼
¡i
(1)
¤
~¢ r
~ £~
aE (l0 ; m 0) fl 0 (kr) = p
jl (kr< ) h l (kr> ) Y lm (µ; Á) Ylm
µ ;Á L
J (~
x0 ) d 3 ~
x0 Y
c
l0 (l0 + 1)
l; m
Z
´
¡ 0 0¢ 0 ³ 0
ik4¼
¤
~¢ r
~ £~
aE (l; m) g l (kr) =
p
h (1)
jl (kr 0) Ylm
µ ;Á L
J (~
x0 ) d 3 ~
x0
l (kr)
c l (l + 1)
So far these expressions are exact.
Now wezd like to express the results in terms of the multipole moments of the source.
20
First we evaluate:
³
´
~ r
~ £~
L¢
J
´ ³
´
1³
~ ¢ r
~ £~
~r £ r
J
i
¢
1¡
=
rj @k @j Jk ¡ rj r2 J j
i
where we used a result from the front cover of Jackson. Now we work on the last term:
rj r2 Jj
and thus
³
´
~
~ £~
L¢ r
J
=
= @k (rj @k Jj ) ¡ (@k rj ) @k J j
= @k (@k rj Jj ¡ (@k rj ) J j ) ¡ ±kj @k J j
³
´
~ ¢~
= r2 ~
J ¢ ~r ¡ @k ± kj Jj ¡ r
J
³
´
~ ¢J
~
= r2 ~
J ¢ ~r ¡ 2r
=
=
=
=
=
³
´
i
1 h³ ~ ´ ~ ~
~ ¢~
~r ¢ r r ¢ J¡r2 ~
J ¢ ~r + 2 r
J
i ·µ
¶
³
´¸
1
@
2 ~
~ ¢ J¡r
~
2+r
r
J ¢ ~r
i
@r
·
³
´
³
´¸
¡1 @
~ ¢J
~ +r2 J
~ ¢ ~r
i
r2 r
r @r
·
³
´¸
¢
¡1 @ ¡
~ ¢ ~r
i
i!½r 2 +r2 J
r @r
³
´
¢
1 @ ¡
!½r2 +ir2 ~
J ¢ ~r
r @r
(1)
and so equation (21) with gl = h l becomes:
µ
Z
³
´¶
¢
ik4¼
1 @ ¡
¤
aE (l; m) = p
jl (kr) Ylm
(µ; Á)
!½r2 +ir2 ~
J ¢ ~r
d3 ~
x
r @r
c l (l + 1)
Now the ®rst term may be rewritten using integration by parts in r :
Z
Z
¢
¯ rmax
1 @ ¡
@
j l (kr)
!½r 2 r2 dr = rj l (kr) !½r2 ¯ 0
¡
(rj l (kr)) !½r2 dr
r @r
@r
where the integrated term is zero provided we take rmax outside the source. Then:
µ
¶
Z
k2 4¼
@
jl (kr) 2 ³~ ´ 3
¤
aE (l; m) = p
Y lm (µ; Á) ¡i½ (rjl (kr)) ¡
r J ¢ ~r
d ~
x
@r
kc
l (l + 1)
Now we work on the last term. From Greenzs theorem
Z
Z µ
¶
¡
¢
@Á
@Ã
Ãr2 Á ¡ Ár2 Ã dV =
Ã
¡Á
dA
@n
@n
~ ¢ ~r and à = jl (kr) Y ¤ (µ; Á) : Then Á = 0 on the surface provided it is
Now we take Á = J
lm
outside the source, so the right hand side is zero. Also,
¤
¤
r2 jl (kr) Ylm
= ¡k2 j l (kr) Y lm
21
and so ®nally we have:
aE (l; m) = p
k2 4¼
l (l + 1)
Z
Ã
@
(µ; Á) ¡i½ (rjl (kr)) +kjl (kr)
@r
¤
Y lm
Ã
~
J ¢ ~r
c
!!
d3 ~
x (22)
Now what about aM : We just need to rearrange
´
³ ³
´
³
´´
1³
1
~
~ ¢~
~ £J
~ = 1 J¢
~ r
~ £ ~r ¡ r¢
~ ~r £ ~
L¢~
J =
~r £ r
J = ~r ¢ r
J
i
i
i
´
³
´
¡1 ~ ³
~
~
~
=
r¢ ~r £ J = i r¢ ~r £ J
i
Z
³
´
¡4¼ik2
¤
~ ~r £ ~
aM (l; m) = p
j l (kr) Y lm
(µ; Á) r¢
J d3 ~
x
c l (l + 1)
These expressions are still exact.
Example: the center-fed linear antenna.
Suppose an antenna of length d has a current distribution
· µ
¶¸
d
I = I0 sin k
¡r
e¡i!t
2
or equivalently,
µ µ
¶¶
~j= I0 sin k d ¡ r
[± (¹ ¡ 1) ¡ ± (¹ + 1)] ^
r
2¼ r2
2
Then from the charge conservation, we get:
µ µ
¶¶
1 ~ ~
¡kI0
d
½=
r¢j =
cos k
¡r
[± (¹ ¡ 1) ¡ ± (¹ + 1)]
i!
2¼i!r 2
2
but since ! = kc; then
½=
(23)
µ µ
¶¶
iI0
d
cos
k
¡
r
[± (¹ ¡ 1) ¡ ± (¹ + 1)]
2¼cr2
2
Now for this antenna, ~r £ ~j = 0; so aM = 0: Then from equation (22):
¡ ¡
¢¢
Z
µ
¶
k2 2I0
[± (¹ ¡ 1) ¡ ± (¹ + 1)]
cos k d2 ¡ r ¡@@r ¡(rj l (kr))
¤
¢¢
aE (l; m) =
p
Ylm (µ; Á)
d3~x
r2
+krjl (kr) sin k r ¡ d2
c l (l + 1)
¡ ¡
¢¢ @
Z
µ
¶
4¼k2 I0
cos k d2 ¡ r ¡@r
¡(rjl (kr))
¢¢
=
p
[Y l0¤ (0) ¡ Y l0¤ (¼)]
dr
+krj l (kr) sin k r ¡ d2
c l (l + 1)
where
Y l0¤ (0) ¡ Y l0¤ (¼) =
=
r
2l + 1
(Pl (1) ¡ Pl (¡1))
4¼
(
r
´
0q
2l + 1 ³
l
1 ¡ (¡1) =
2l+1
4¼
¼
22
if
l is even
if
l is odd
Next we rearrange the ®rst term in the integrand:
µ µ
¶¶
·
µ µ
¶¶
¸
µ µ
¶¶
d
@
@
d
d
cos k
¡r
(rj l (kr)) =
cos k
¡r
rjl (kr) ¡k sin k
¡r
rjl (kr)
2
@r
@r
2
2
Thus
aE (l; 0) =
=
=
=
p
·
µ µ
¶¶
¸
Z
4k2 I0 (2l + 1) ¼
@
d
p
cos k
¡r
rj l (kr) dr
@r
2
c l (l + 1)
p
¯d=2
µ µ
¶¶
¯
4k2 I0 (2l + 1) ¼
d
p
cos k
¡r
rjl (kr)¯¯
2
c l (l + 1)
0
p
µ ¶
2
4k I0 (2l + 1) ¼ d
kd
p
jl
2
2
c l (l + 1)
s
µ ¶2 µ ¶
8I0 (2l + 1) ¼ kd
kd
jl
cd
l (l + 1)
2
2
where l is odd. Thus there are multipoles of all odd orders.
The full wave (kd = 2¼) and half
p wave (kd = ¼) are of particular interest. For these we
®nd the coef®cients cdaE (l; 0) =8 ¼I0 to be:
l fullpwave
half wave
p
¡
¢
1
2 j (¼) = 1 6¼ 2 sin ¼ ¡ cos ¼
p
p ¼2 ¡¼¢
p ¼2 ¡ 4 ¢
6¼
1
1
6
2
2
¼2
¼
p 1
1
¼
2 6 4 j1 2 = 2 6 4
¼2 = 8 = 0: 305
=
6
=
3:
848
2
p
p
¡ ¢
1
3 16 p21¼ 2 j3 (¼) = 1: 2473
21¼ 2 j l ¼2 ¡ =¢ 6: 0544 £ 10 ¡2
24 p
1
2
1
5 30
330¼ 2 j5 ¡ ¼2 ¢ = 3: 1233 £ 10 ¡4
480 p
p330¼ 2 j5 (¼) = 0: 11914
1
¡3
1
2
¼
¡3
7 28 210¼ j7 (¼) = 5: 6673 £ 10
112 210¼ j7 2 = 2: 0037 £ 10
~ l0 (equation 14)
Finally we need the X
µ
¶
¡1
1p
1p
~
Xl0 = p
l (l + 1)Yl;1 (^
x¡i^
y) +
l (l + 1)Y l;¡ 1 (^
x+i^
y)
2
l (l + 1) 2
s
Ãs
!
r
¡1 2l + 1
(l ¡ 1)! 1 iÁ
(l + 1)! ¡ 1 ¡iÁ
=
P e (^
x¡i^
y) +
P e
(^
x + i^
y)
2
4¼
(l + 1)! l
(l ¡ 1)! l
s
r
¡
¢
¡1 2l + 1 (l ¡ 1)!
d
=
(¡1) sin µ P l (¹) eiÁ (^
x¡i^
y) ¡ e¡iÁ (^
x + i^
y)
2
4¼
(l + 1)!
d¹
s
2l + 1 (l ¡ 1)!
d
= i
sin µ Pl (¹) (sin Á^
x¡ cos Á^
y)
4¼ (l + 1)!
d¹
s
2l + 1 (l ¡ 1)!
d
^
= ¡i
sin µ P l (¹) Á
4¼ (l + 1)!
d¹
and thus
~ l0 = i
^r £ X
s
2l + 1 (l ¡ 1)!
d
sin µ P l (¹) µ^
4¼ (l + 1)!
d¹
23
and
~ l0
aE (l; 0) ^
r£ X
=
=
µ ¶2 µ ¶ s
(2l + 1) ¼ kd
kd
2l + 1 (l ¡ 1)!
d
jl
i
sin µ P l (¹) ^
µ
l (l + 1)
2
2
4¼ (l + 1)!
d¹
µ ¶2 µ ¶
4I0 (2l + 1) kd
kd
d
i
jl
sin µ P l (¹) µ^
cd l (l + 1) 2
2
d¹
8I0
cd
s
Thus the power is:
<
=
=
=
¯
¯
h
i X 0 h
i¤¯
dP
c ¯¯X
l+ 1
l
+1
0
~ l0 ¢
~ l 00 ¯¯
>=
(¡i)
¡aE (l) ^
r £X
i
¡aE (l ) ^r £ X
¯
d8¼k 2 ¯
¯
l odd
l0
¯
¯
µ
¶
µ
¶
µ
2
X same with l ! l0 ¶¯¯
c ¯¯ X
kd
kd
d
l 4I 0 (2l + 1)
(¡i) i
jl
sin µ Pl (¹) ¢
¯
¯
and i ! ¡i
8¼k2 ¯
cd l (l + 1)
2
2
d¹
¯
l odd
l0
¯
¯
µ
¶ 2 µ ¶4
µ
¶
µ
¶
¯X
¯
l¡1 (2l + 1)
c
4I0
kd
kd d
same
¯
¯
sin 2 µ ¯
(¡1) 2
jl
P l (¹)
¯
0
2
with l ! l
8¼k
cd
2
¯
l (l + 1)
2
d¹
¯
l
¯
¯
µ ¶2
µ ¶
¯X
¯
l¡1
I02
kd
2l + 1
kd d
¯
¯
sin2 µ ¯
(¡1) 2
jl
P l (¹) (same with l ! l0 )¯
2¼c 2
¯
l (l + 1)
2
d¹
¯
l
Thus for the full wave antenna, we get:
¯Ã
¯
¡1 ¡ 3
¢¢ !
¯
¯
d
7
d
(¹) ¡ 12
j 3 (¼) d¹
5¹
¡
3¹
dP
I02 2 2 ¯ 32 j 1 (¼) d¹
¯
¡ 63 5 35 23 15 ¢
<
>=
¼ sin µ ¯
(same)¯
11
d
¯
+ 30 j 5 (¼) d¹ 8 ¹ ¡ 4 ¹ + 8 ¹
¯
d2¼c
µ
µ
¶¶
2
¡
¢ 11
I02 2
3
7
21 4
1
=
¼ sin 2 µ
j 1 (¼) ¡ j 3 (¼) 5¹2 ¡ 1 +
j 5 (¼)
¹ ¡ 7¹2 +
2¼c
2
8
8
2
2
µ
¶
µ
µ
¶¶2
2
¢ 11 j5 (¼) 21 4
I02 2
3
7 j 3 (¼) ¡ 2
1
=
¼ sin 2 µ
j 1 (¼)
1¡
5¹ ¡ 1 +
¹ ¡ 7¹2 +
2¼c
2
12 j 1 (¼)
12 j1 (¼) 2
2
But
sin ¼
cos ¼
1
j 1 (¼) =
¡
= = 0:318 31
2
¼
¼
¼
so
µ
¶
¢ 11 J5:5 (¼) ¡
¢ 2
dP
I2 9
7 J3: 5 (¼) ¡ 2
<
>= 0
sin2 µ 1 ¡
5¹ ¡ 1 +
21¹4 ¡ 14¹2 + 1
d2¼c 4
12 J1: 5 (¼)
12 J1:5 (¼)
2
¡
¡
¢
¡
¢¢2
I0 9
=
sin 2 µ 1 ¡ 0:303 23 5¹2 ¡ 1 + 5: 741 0 £ 10¡ 2 21¹4 ¡ 14¹2 + 1
2¼c 4
¡
¢2
9 I02
=
sin 2 µ 1:360 6 ¡ 2:319 9¹2 + 1:205 6¹4
4 2¼c
¡
¢
9 I02
=
sin 2 µ 1:8125 ¡ 5:7054¹2 + 6:9246¹4 ¡ 3:8319¹6 + 0:81757¹8
4 2¼c
Jacksonzs exact result (9.57) is the blue-green line
¡
¢
dP
I2 4 cos4 ¼2 cos µ
<
>=
d2¼c
sin2 µ
24
Two term result
³ (red line) : ³
´´2
2
2
9
(sin
µ)
1
¡
0:
30323
5
(cos
µ)
¡
1
4
the 3 term ¡result is almost the same (black line)
¢
2
9
(sin µ) 1: 8125 ¡ 5: 7054 cos2 µ + 6: 9246 cos 4 µ ¡ 3: 8319 cos 6 µ + : 81757 cos8 µ
4
4
2
0
-2
-4
The blue dashed line is the 1-term (dipole) result.
Half-wave antenna:
¯Ã
¯
dP
I02 ¼ 2
2 ¯
<
>=
sin µ ¯
d2¼c 4
¯
3
j
2 1
Here
¡¼¢
j1
so
<
dP
I2 9 ¼ 2
>= 0
d2¼c 4 4
µ
4
¼2
¡ ¢
d
7
(¹) ¡ 12
j3 ¼2
d¹ ¡ ¢
¡
+ 11
j ¼ d 63
¹5
30 5 2 d¹
8
2
¶2
2
sin µ
³¼ ´
Ã
2
=
4
¼2
¡ ¡
d
1
d¹ 2
¡ 35
¹3
4
5¹3 ¡ 3¹
¢
+ 15
¹
8
¢¢ !
¯
¯
¯
(same)¯
¯
!2
¡ ¢
¡ ¢
¢ 11 J 5:5 ¼2 ¡
¢
7 J3:5 ¼2 ¡ 2
4
2
¡ ¢ 5¹ ¡ 1 +
¡ ¢ 21¹ ¡ 14¹ + 1
1¡
12 J1:5 ¼2
12 J 1:5 ¼2
¡
¡
¢
¡
¢¢2
I02 9
sin2 µ 1 ¡ 4: 624 1 £ 10 ¡2 5¹2 ¡ 1 + 1: 891 2 £ 10 ¡3 21¹4 ¡ 14¹2 + 1
2
2¼c ¼
¡
¢2
I02 9
=
sin2 µ 1: 048 1 ¡ 0:257 68¹2 + 3: 971 5 £ 10¡ 2 ¹4
2
2¼c ¼
¡
¢2
(sin µ)2 ¼92 1: 048 1 ¡ 0:257 68 cos2 µ + 3: 971 5 £ 10 ¡2 cos4 µ
¡
¡
¢¢2
(sin µ)2 ¼92 1 ¡ 4: 624 1 £ 10¡2 5 cos2 µ ¡ 1
=
25
Jacksonzs exact result (9.57) is the blue-green line
¡
¢
dP
I 2 cos 2 ¼2 cos µ
<
>=
d2¼c
sin2 µ
1
0.8
0.6
0.4
0.2
-0.20
-0.2
0.2
-0.4
-0.6
-0.8
-1
Here the two-term (red dotted) and 3-term(black) results are indistinguishable. Again
the dashed blue
line is the 1-term (dipole) result.
Now we specialize to the case in which the source size d ¿ ¸: This means that kr ¿ 1
in each integrand, and we may use the small argument expansion for the Bessel function jl :
l
jl (kr) =
(kr)
(2l + 1)!!
and
@
@ kl rl+1
l+1
l
(rjl (kr)) =
=
(kr)
@r
@r (2l + 1)!!
(2l + 1)!!
The quantity that appears in the integrand of aE (equation 22) is
Ã
!
~
@
J ¢ ~r
l+1
k (kr)l ~
¡i½ (rjl (kr)) +kjl (kr)
= ¡i½
(kr)l +
J ¢ ~r
@r
c
(2l + 1)!!
c (2l + 1)!!
Ã
!
l
~
(kr)
J ¢^
r
=
¡i½ (l + 1) + kr
(2l + 1)!!
c
26
The second term is smaller than the ®rst by a factor kr; and so we drop it. Then:
Z
¡ik2 4¼
l+1
l
¤
aE (l; m) = p
Ylm
(µ; Á) ½ (kr) d3 x
~
l (l + 1) (2l + 1)!!
r
¡4¼ik l+2 l + 1
=
Q lm
(2l + 1)!!
l
where
Z
¤
Q lm = ½r l Ylm
(µ; Á) d3 ~
x
is the spherical multipole moment of the source.
Similarly,
r
4¼ikl+ 2
l+1
aM (l; m) =
M lm
(2l + 1)!!
l
where
Ã
!
Z
~
¡1
J
¤
~ ~
Mlm =
r¢
r£
rl Y lm
(µ; Á)
l+1
c
is the magnetic multipole.
12 Atomic systems
The transition probability is de®ned in terms of the radiated power by
P
1
= = transition probability
h!
¿
where ¿ is the mean lifetime of the state. The charge density is assumed to have the form
½ 3e
Y (µ; Á) f or r < a
a 3 lm
½ (~
x) =
0
f or r > a
So that
Ql 0m 0
=
=
=
3e
a3
Z
a
r l r2 dr
0
l+3
Z
Y lm (µ; Á) Y l¤0 m0 (µ; Á) d-
3e a
± ll 0 ±mm0
a3 l + 3
3e l
a ± ll0 ±mm0
l +3
and thus
¡4¼ikl+2
aE (l; m) =
(2l + 1)!!
27
r
l + 1 3e l
a
l l+3
Then from equation (18) for the power, we have
µ
¶2
2
1
c
3e l
(4¼) k2l+4 l + 1
'
a
¿
8¼k2 h! l + 3
[(2l + 1)!!]2 l
µ
¶
µ
¶
2
e2
3
l+1
2¼
2l
=
2 (ka) !
hc l + 3
l
[(2l + 1)!!]
2l
Because of the factor (ka) ; the transition probability falls off rapidly with increasing
l;and so the lowest non-vanishing multipole is usually the only one we have to consider.
¡ Z ¢2
The magnetic multipole contribution is smaller by a factor of order 137
where Z is the
effective nuclear charge (i.e. the charge interior to the transitioning electron.).
13 Scattering of EM waves by a spherical object
13.1
Plane wave expansion
Suppose an incoming plane wave is scattered by a spherical object. We expect the outgoing
waves to be described in terms of vector spherical harmonics. Thus our ®rst task is to
expand the incoming wave in terms of the eigenfunctions (8).
We have already found the Greenzs function, and we may expand it in the limit r 0 À r :
0
0
eikR
eik j~x¡x~ j
e ikr ¡ ik¢~
~x
¯
¯
=
'
e
4¼R
4¼r 0
4¼ ¯ ~
x¡~
x0 ¯
(Note: this is just a mathematical manipulation. We are not yet attaching any physical
meaning to ~
x and ~
x0 .) The Greenzs function expansion (2) may be evaluated with r> = r0 ;
and using the large argument expansion for the Bessel functions in kr0 :
0
X
¡ 0 0¢
eikr ¡ i~k¢~x
(1)
¤
e
= ik
j l (kr) h l (kr0 ) Y lm (µ; Á) Ylm
µ ;Á
0
4¼r
l;m
=
ik
X
l+1
j l (kr) (¡i)
l;m
0
0
¡ 0 0¢
eikr
¤
Ylm (µ; Á) Y lm
µ ;Á
0
kr
and the eikr =r0 term may be brought out of the sum since it does not depend on l or m:
Then:
X
¡ 0 0¢
~
¤
e¡ik¢~x = 4¼
jl (kr) (¡i)l Y lm (µ; Á) Ylm
µ ;Á
l; m
and taking the complex conjugate, we have
X
¡
¢
~
l
¤
e ik¢~x = 4¼
jl (kr) (i) Y lm
(µ; Á) Y lm µ 0; Á0
l; m
Now only the terms in Y lm depend on m: Using the addition theorem (Jackson equation
28
3.62),
X
l;m
¡
¢ X
2l + 1
¤
Y lm
(µ; Á) Ylm µ 0 ; Á0 =
P l (cos °)
4¼
l
where ° is the angle between ~
x and ~
x0; or equivalently, between ~
x and ~
k: Finally then:
X
~
eik¢~x =
i l (2l + 1) jl (kr) Pl (cos °)
l
=
X p
i l 4¼ (2l + 1)j l (kr) Y l0 (°)
l
This is exactly the factor that appears in the plane wave expression.
Now we also need to worry about the polarization of the incoming wave. Any wave can
be decomposed into either linear or circular polarizations. The circular polarizations turn
out to be more convenient here. So write:
~ § = (^
E
"1 § i^"2 ) eikz
(24)
where now we have chosen the z¡axis along ~
k: Then
1
¹ §= r
~ £E
~ § = ^"3 £ E
~§
B
ik
Now we write each ®eld as an expansion in the eigenfunctions:
X
¡
¢
~§ =
¹ lm + i b § (l; m) r£
~
¹ lm
E
a§ (l; m) j l X
jlX
k
(25)
l;m
and
~§ =
B
X
l;m
¡
¢
¹ lm ¡ i a§ (l; m) r£
~
¹ lm
b § (l; m) j l X
jlX
k
~ lm ;
where, by the orthogonality of the X
a§ (l; m) jl =
Z
~¤ ¢ E
~ § dX
lm
(26)
(We have already proved most of the orthogonality relations we need here in section 8.
To show that
¡
¢
¹ lm ¢ r£
~
¹ lm = 0
X
jl X
~ l is perpendicular to X
~ lm ; and the remaining term involves
expand the curl. The term in rj
³
´
¹ lm ¢ r
~ £X
¹ lm = X
¹ lm ¢ ~
~ Y lm = ¡X
¹ lm ¢ rY
~ lm
X
r r2 ¡ r
¹ lm / ~r £ rY
~ lm is perpendicular to rY
~ lm ; and so the relation
(equation 9 ) . But X
follows.)
Similarly:
Z
~¤ ¢ B
~ § db§ (l; m) jl =
X
(27)
lm
29
Now insert the value (24) into (26):
Z
~ ¤ ¢ (^
a§ (l; m) j l =
X
"1 § i^"2 ) eikz dlm
Z
1
¤ ikz
= p
L§ Ylm
e dl (l + 1)
p
Z
l (l + 1) ¡ m (m § 1)
¤
ikz
=
p
Ylm§
d1e
l (l + 1)
p
Z
X 0p
l (l + 1) ¡ m (m § 1)
¤
=
p
Ylm§
i l 4¼ (2l 0 + 1)j l0 (kr) Y l0 0 (µ) d1
l (l + 1)
l0
Now by orthogonality of the Y lm;only l = l0 and m = ¨1 contribute:
p
l (l + 1) ¡ m (0) l p
a§ (l; m) j l =
p
i 4¼ (2l + 1)j l (kr) ± m;¨1
l (l + 1)
p
a§ (l; m) = il 4¼ (2l + 1)±m;¨1
(28)
0
with a similar result for the b s. Since m = §1; the lowest allowable value of l is 1. Thus
we can expand the plane wave ®elds as:
·
1
´¸
X
p
1~ ³
l
~
~
~
E§ (~
x) =
i 4¼ (2l + 1) jl (kr) Xl;§1 § r£ j l (kr) X l;§1
(29)
k
l=1
where, as we showed in section 10, m = § corresponds to §1 unit of angular momentum in
~ is:
the z¡direction. The corresponding result for B
·
1
´¸
X
p
i ~ ³
l
~
~
~
B§ (~
x) =
i 4¼ (2l + 1) ¨j l (kr) X l;§ 1 ¡ r£ jl (kr) Xl;§1
k
l=1
13.2
Application to scattering
The total electric ®eld is a superposition of the incoming and outgoing ®elds:
~ total = E
~ incident + E
~ scattere d
E
The scattered waves are ougoing at in®nity and so are represented with h( 1) :
³
´
X
i
(1) ~
(1) ~
~ sc attered =
~
E
a (l; m) h l X
b (l; m) r£
hl X
lm +
lm
k
l;m
Now if the scatterer is a conducting sphere, the total tangential electric ®eld on the surface
of the sphere must be zero. We have already expanded the curl several times. The result is:
³
´
@h
(1) ~
~
~ lm + hr
~ £X
~ lm
r£
hl X
=
^r £ X
lm
@r
³³
´
´
@h
~ lm + h
~ Y lm
=
^r £ X
r r2 ¡ r
~
@r
i
30
~ which is found from:
We only need the tangential part of r;
³
´
@
~ =~
~
^r£ ~r £ r
r
¡ rr
@r
and thus
~ = ¡i ^r £ L
~
r
r
and so:
³
´ @h
r
(1) ~
~
~ lm + h ^
~ lm + h ~rr2 Y lm
r£
hl X
r £X
^
£X
lm =
@r
r
i
~ is:
and thus the tangential part of E
Ã
!
(1)
(1)
X
@h
h
i
(
1)
l
l
~ scatte red =
~ lm + b (l; m)
~ lm
E
a (l; m) hl X
+
r£ X
^
k
@r
r
l;m
At r = a; the surface of the sphere:
·
¯
¸
1
X
p
¯
1 @
l
~
~
¯
~
Etan = E 0
i 4¼ (2l + 1) jl (ka) X l;§ 1 §
(rj l (kr))¯
^r £ X l;§ 1
ka @r
r=a
l=1
¯
X
i
@ ³ (1) ´¯¯
( 1)
~
~ lm
+
a (l; m) hl (ka) Xlm +
b (l; m)
rh l ¯
^r £ X
ka
@r
r=a
l;m
= 0
Thus a (l; m) and b (l; m) = 0 unless m = §1: We can simplify the results by writing:
p
E0
a (l; m) = il 4¼ (2l + 1)
®lm
2
and
p
E0
b (l; m) = i l¡ 1 4¼ (2l + 1) ¯ lm
2
~ tan = 0 requires:
Then E
® l;§ 1 (1)
jl (ka) +
h l (ka) = 0
2
and thus:
®l;§1
= ¡2
( 1)
jl (ka)
( 1)
hl
(ka)
=¡
hl
(2)
(ka) + h l (ka)
(1)
h l (ka)
(2)
= ¡1 ¡
and similarly
§2
h l (ka)
(1)
h l (ka)
¯
¯
¯
@
@ ³ (1) ´¯¯
(rjl (kr))¯¯
+ ¯ l;§1
rh l ¯
=0
@r
@r
r= a
r= a
31
So
¯ l;§1
=
=
¨
@
@r
¨1 ¨
³
(1)
rh l
@
@r
@
@r
@
@r
´
³
+
@
@r
´
(1)
rhl
³
´¯
(2) ¯
rhl
¯
³
´ ¯¯
(1)
¯
rhl
³
(2)
rh l
´¯
¯
¯
¯
¯
¯
r=a
r =a
We can calculate the radiated power using equation (19). For each polarization separately,
we have:
¯
¯
³
´¯ 2
p
dP §
cE 02 ¯¯ X
¯
~ l;§ 1 + ¯ ^
~
=
(¡i)l+1 i l ¼ (2l + 1) ®l X
¯
l§ r £ Xl;§1 ¯
2
¯
¯
d8¼k
l
and thus the scattering cross section is:
¯
¯
³
´¯2
d¾ §
1 ¯¯ X p
~ l; §1 + ¯ l§^r £ X
~ l;§ 1 ¯¯
=
¡i ¼ (2l + 1) ®l X
¯
d2k 2 ¯
¯
l
p
(Note that the magnitude of the incident ®eld is 2E 0 ):
Now we can compute the ®elds in the long-wavelength limit. For ka ¿ 1; h l ' inl ; and
we have:
®l
=
¡2
l
j l (ka)
(1)
h l (ka)
= ¡2
2l+1
l+1
(ka)
(ka)
(2l + 1)!! (¡i (2l ¡ 1)!!)
2
(ka)
i (2l + 1)!! (2l ¡ 1)!!
So clearly l = 1 is the dominant term. Similarly for ¯ l;§1 we have:
¯
¡ l l+1 ¢¯
@
@
¯
(rj l (kr))¯ r=a
2 @r
k r
kl+1
@r
r=a
¯
¯
¯ l;§1 = ¨2
=
§
i @ (rnl )¯
i (2l + 1)!! (2l ¡ 1)!! @ (r¡l )¯
=
@r
r =a
@r
2l+1
=
For l = 1 :
2
(l + 1) (ka)
l+1
§
=¨
®l
i (2l + 1)!! (2l ¡ 1)!! (¡l)
l
and
Then to lowest order:
2
3
® 1 = ¡ i (ka)
3
4
3
¯ 1; §1 = § i (ka)
3
¯
¯2
d¾ §
1 4
6
¯~
¯
~
=
(ka)
3¼
X
§
2i^
r
£
X
¯
1;§1
1;§1 ¯
d2k2 9
32
r=a
¯
¯2
¯~
¯
We have already calculated ¯X
1;§ 1 ¯ =
¢
1 + cos2 µ : But here we have a cross term:
³
´¤
~ 1;§1 ¢ ^r £ X
~ 1;§1
X
3
16¼
¡
From equation (14), we have
µ
µ
¶ p
µ
¶
¶
¡1 p
x¡i^
^
y
x + i^
^
y
~ 1;§1 = p
X
2 ¨ 1 (§1 + 1)Y1;§1+1
+ 2 ¨ 1 (§1 ¡ 1)Y 1;§ 1¡1
§ Y1§1^z
2
2
2
So:
µ
µ
¶
¶
¡1 p
x + i^
^
y
~ 1;1 = p
X
2Y l;0
+ Y 1;1 ^
z
2
2
Ãp r
!
r
¡1
2
3
3
iÁ
= p
cos µ (^
x + i^
y) ¡
sin µe ^z
2
4¼
8¼
2
r
¡1
3 iÁ
= p
e (cos µ (cos Á ¡ i sin Á) (^
x + i^
y) ¡ sin µ^z)
2 8¼
r
1 3 iÁ
= ¡
e (cos µ (cos Á^
x + i cos Á^
y ¡ i sin Á^
x + sin Á^
y ) ¡ sin µ^
z)
4 ¼
r
´
1 3 iÁ ³
^¡^
= ¡
e
i cos µ Á
µ
4 ¼
and
Ãp r
!
r
¡1
2
3
3
¡iÁ
~ 1; ¡1 = p
X
cos µ (^
x¡i^
y) +
sin µe ^
z
2
4¼
8¼
2
r
´
1 3 ¡ iÁ ³
^+^
=
e
i cos µ Á
µ
4 ¼
So
r
´
1 3 § iÁ ³^
~
^
X1;§1 =
e
µ¨i cos µ Á
4 ¼
Then
r
´
1
3 § iÁ ³ ^
~ 1;§1 =
^
r £X
^
e
Á§i cos µ µ
4 ¼
And so the cross terms are of the form
³
´¤
~ 1;§1 ¢ ^r £ X
~ 1;§ 1 = ¨i 3 cos µ
X
8¼
and so
³
´ ³
´¤
d¾ §
2 2
4 ~
~ 1;§1 ¢ X
~ 1; §1 § 2i^
~ 1;§1
=
¼a (ka) X
r £X
r £X
1;§1 § 2i^
d3
µ
¶
¢
2 2
3 ¡
3
4
=
¼a (ka) 5
1 + cos 2 µ ¡ 2
cos µ
3
16¼
4¼
µ
¶
¡
¢
5
4
= a2 (ka)
1 + cos2 µ ¡ cos µ
8
33
which agrees with equation 9.94.
This result shows explicitly the interference between the modes.
1
0.5
-2
-1.5
-1
-0.5
0
-0.5
-1
scattering cross section versus angle
34