Perturbation theory for scattering in an almost
uniform medium
S.M.Lea
Suppose we have a medium with average dielectic constant "0 and magnetic permeability
¹0 ; but with " 6= " 0 and ¹ 6= ¹0 in small regions. Note: We are in Gaussian units. "0 is
NOT the SI " 0 ; but the unperturbed value of ":
Maxwellzs equations for this system are:
~ ¢B
~ =0
r
(1)
(no free charge)
and
together with
~ ¢D
~ =0
r
(2)
~
~ £E
~ = ¡ 1 @B
r
c @t
(3)
~
~ £H
~ = 1 @D
r
c @t
(4)
~ = "E
~
D
(5)
and
~ = ¹H
~
B
(6)
We want to separate out those regions which deviate from the uniform properties, so we
write:
~ =D
~ ¡ "0 E
~ + "0 E
~
D
Then
Ã
!
h
³
´i
³
´
~
1 @B
~
~
~
~
~
~
~
~
r £ r £ D ¡ "0 E = r £ r £ D ¡ " 0 r £ ¡
c @t
where we used equation (3). Thus:
h
³
´i
³
´
³
´
~ £ r
~ £ D
~ ¡ "0 E
~
~ r
~ ¢D
~ ¡ r2 D
~ + "0 @ r
~ £B
~
r
= r
c @t
´
"0 @ ³ ~
2~
~
= ¡r D+
r£B
c @t
~ :
where we used equation (2). Now we do the same kind of thing with B
~ =B
~¡¹ H
~ +¹ H
~
B
0
0
1
so
~ £B
~
r
=
=
³
´
~ £ B
~ ¡¹ H
~ +¹ H
~
r
0
0
³
´
~
~ £ B
~ ¡¹ H
~ + ¹0 @ D
r
0
c @t
(equation 4) Thus:
"
#
h
³
´i
³
´ ¹ @D
~
"
@
0
2
0
~ £ r
~ £ D
~ ¡ "0 E
~
~ +
~ £ B
~ ¡¹ H
~ +
r
= ¡r D
r
0
c @t
c @t
and thus:
h
³
´i " @
³
´
~
¹0 "0 @ 2 D
0
~ £ r
~ £ D
~ ¡ "0 E
~
~ £ B
~¡¹ H
~
=
¡
r
+
r
(7)
0
c2 @t2
c @t
~ and its average value
This equation shows that the differences between the local value of D
~ ave = "0 E
~ ; and between the local value of B
~ and its average value B
~ ave = ¹0 H
~ act as
D
~
sources for D:
Now let all quantities have the oscillating form e¡i!t : The equation (7) becomes
h
³
´i
³
´
¡ 2
¢
~ = ¡r
~ £ r
~ £ D
~ ¡ "0E
~ ¡ i "0 ! r
~ £ B
~ ¡¹ H
~
r + k2 D
0
c
where
¹ "0
!2
k2 = 02 ! 2 = 2
c
v
p
and v = c= ¹0 "0 is the wave speed in the medium. We can use the Greenzs function for the
~ :
wave equation to obtain an expression for D
Z
h
³
´i
³
´o
exp (ik j~x ¡ ~x0 j) n ~
"0! ~
~ =D
~ (0)+ 1
~ £ D
~ ¡ "0 E
~
~ ¡¹ H
~
D
dV
r
£
r
+
i
r
£
B
0
4¼
j~x ¡ ~x0 j
c
(8)
~ appears on the right hand side as well.
Of course this is not really a solution yet, because D
Now we expect the solution to be a radiation ®eld of the form
~ ¡
r 2D
ikr
~ =D
~ ( 0) + A
~ sc e
D
r
where, from equation 8, the scattering amplitude is:
Z
n 0 h 0 ³
´i
´o
"0 ! ~ 0 ³ ~
~ sc = 1
~ £ r
~ £ D
~ ¡ "0 E
~
~
A
dV 0 exp (ik^
r ¢ ~x 0 ) r
+i
r £ B ¡ ¹0 H
4¼
c
exp ( ik j ~
x¡~
x0 j)
(We have made the usual approximations in evaluating
): Now we do the
j~
x¡~
x0 j
}integration by parts} trick. The integral is of the form:
Z
Z
Z
³
´
³
´
³
´
~ £ r
~ £ ~y dV =
~ £ eik^r ¢~x r
~ £ ~y dV ¡ re
~ £ r
~ £ ~y dV
exp (ik^r ¢ ~x) r
r
Z
Z
³
´
³
´
~ £ ~y dS ¡ ik^r £ eik^r ¢~x r
~ £ ~y dV
=
n
^ £ eik^r¢~x r
Z
³
´
~ £ ~y dV
= 0 ¡ ik^r £ eik^r¢~x r
2
Then we can do it again to get:
µ Z
¶
(ik)2 r^ £ r^£ eik^r¢~x ~y dV
~ is
Thus the expression for A
·
³ ³
´´ r "
³
´¸
2 Z
k
0
ik^
r¢~
x0
~
~
~
~
~
Asc =
dV e
¡^r£ ^
r £ D ¡ "0 E +
^r£ B ¡ ¹0 H
4¼
¹0
Z
·
³ ³
´´ r "
³
´¸
k2
0
0
~ ¡ "0 E
~
~¡¹ H
~
=
dV eik^r¢~x ¡^r£ ^
r£ D
+
^r£ B
0
4¼
¹0
(Jackson and I differ by a sign here.)
In this expression we can recognize the electric dipole moment (pg 4 of your multipole
notes)
Z
³
´
0
~ ¡ "0 E
~ dV
~p =
eik^r¢~x D
and the magnetic dipole moment (pg 7)
r Z
³
´
"0
0
~ ¡¹ H
~ dV
m
~ =¡
eik^r¢~x B
0
¹0
The differential cross section for scattering into polarization ^" is:
¯
¯
¯ ¤ ~ ¯2
"
^
¢
A
¯
¯
sc
d¾
=
¯
¯2
d¯ ~ ( 0) ¯
¯D ¯
=
k4
¤
r £ ^"¤ ) ¢ m
~j
¯
¯ j^" ¢ ~p + (^
¯ ~ (0)¯ 2
¯D ¯
Now to get an explicit solution we must approximate. We write
~ = "0 E
~ + ±"E
~
D
~ Then
and similarly for B:
~ ¡ "0E
~
D
=
~ =
±"E
±" ~ (0)
D
"0
±"
D0 ^
e0 eik^n0¢~x
"0
where we can use the unperturbed ®eld on the right hand side since it is multiplied by the
small quantity ±"
"0 : Then the dipole moment is
Z
0 ±"
n 0¢~
x
~p =
eik^r¢~x
dV D0e^0 eik^
"0
=
3
Similarly
r
Z
0 ±¹
"0
¡
eik^r¢~x
dV B 0 ^
n0 £ ^
e0
¹0
¹0
Z
0 ±¹
= ¡ eik^r¢~x
dV D0 ^
n 0 £ ^e0
¹0
!p
~0 = i! B
~
~ 0 = "0 ! B
~
since, from Faradayzs law, ik^
n0 £ E
^0 £ D
c 0 ; and so c ¹0 "0 n
c 0;
q
¹0
B0 =
D0: Then the scattering cross section is:
"0
¯
·
¸¯ 2
Z
¯ 1
¯
d¾
±"
±¹
= k4 ¯¯
dV 0 exp (i~q ¢ ~x 0 )
e0 ¢ ~
^
"¤ ¡
(^r £ ^
"¤ ) ¢ (^
n0 £ ^
e0 ) ¯¯
(9)
d4¼
"0
¹0
where
~q = k (^
n 0 ¡ ^r)
If the wavelength ¸ À the scale of the -uctuations ±" and ±¹; then ei~q¢~x ¼ 1; and we
have dipole radiation.
You will need equation (9) in problem 10.20.
m
~
=
4